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Title: 2a notes for 12th standard
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Description: its very to read and understand
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BINOMIAL THEOREM
*
Binomial Theorem for integral index:
n
n
n
n
If n is a positive integer then (x + a)n = C0 xn + C1 xn-1 a + C2 xn-2 a2 + …
...
+ Cn an
...
*
In the expansion, the sum of the powers of x and a in each term is equal to n
...
C2 … Cn are called binomial coefficients and
these are simply denoted by C0, C1, C2 …
...
n
*
C0 = 1, nCn = 1, nC1 = n, nCr = nCn − r
In the expansion, (r+1)th term is called the general term
...
Thus Tr+1 = Cr xn-r ar
...
n
* (x – a)n =
∑ nC r
r =0
n
xn-r (-a)r =
∑
r= 0
n
(-1)n Cr xn-rar
n
n
n
n
= C0 xn - C1 xn-1 a + C2 xn-2a2 - …
...
sakshieducation
...
sakshieducation
...
*
n
+ 1
i) If n is even, then 2 th term is the middle term
n +1
n+3
th
ii) If n is odd, then 2
and 2 th terms are the middle terms
...
(n + 1) x
ii) If
x +1
= p + F where p is a positive integer and 0 < F < 1 then (p+1) th term is the
numerically greatest term in the expansion of (1 + x)n
...
iv) (1 – x)-2 = 1 + 2x + 3x2 + 4x3 + … + (r + 1)xr + …
...
n(n − 1)
n(n − 1)(n − 2)
2
3!
vi) (1 – x) = 1 + nx + 2! x +
x3
+ …
...
sakshieducation
...
sakshieducation
...
( n + 2)
C3 x3 + …
...
∞
When x < 1,
2
p x
p( p + q) x
2! q (1 + x)-p/q = 1 – 1! q +
3
p ( p + q)( p + 2q) x
3!
q + ……
...
x n a 0 + nC1
...
x n−2 a 2 +
...
x n−r a r +
...
x 0 a n
Proof:
We prove this theorem by using the principle of mathematical induction (on n)
...
That is
( x + a) k = k C0
...
a 0 + k C1
...
a1 + k C2
...
a 2 +
...
x k −r
...
+ k Ck
...
a k
Now we prove that the theorem is true when n = k + 1 also
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( x + a ) k +1 = ( x + a )( x + a )k
= ( x + a)( k C0
...
a 0 + k C1
...
a1 + k C2
...
a 2 +
...
x k −r
...
+ k Ck
...
a k )
= k C0
...
a 0 + k C1
...
a1 + k C2
...
a 2 +
...
x k −r +1
...
+ k Ck
...
a k + k C0
...
a1 + k C1
...
a 2 +
...
x k −r +1
...
+ k Ck −1
...
a k + k Ck
...
a k +1
k
k
k − r +1 r
...
+ ( k Ck + k Ck −1 )
...
a k
= k C0
...
a 0 + ( k C1 + k C0 )
...
a1 + ( k C2 + k C1 )
...
a 2 +
...
x
+ k Ck
...
a k +1
Since k C0 = 1 =
k +1
C0 ,k Cr + k Cr −1 = ( k +1)Cr for 1 ≤ r ≤ k , k Ck = 1 =
( k +1)
C( k +1)
(x + a)k+1
=
( k +1)
( k +1)
C0
...
a 0 +
Ck
...
a k +
( k +1)
C1
...
a1 +
( k +1)
C2
...
a 2 +
...
x k −r +1
...
+
k +1
Ck +1
...
a k +1
Therefore the theorem is true for n = k +1
Hence, by mathematical induction, it follows that the theorem is true of all positive integer n
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Very Short Answer Questions
1
...
(i) (4x + 5y)
2p
7
2
(ii) x + y
4
3
7
3p
5
6
(iii) −
7
5
(iv) (3 + x – x2)4
i) (4x + 5y)7
Sol
...
x + y =
4
3
5
5
4
3
2
2
3
1
4
2
2 7
2 7
2 7
2 7
7
C0 x + 5C1 x y + 5 C 2 x y + 5C3 x y + 5C4 x y + 5C5 y
3
3 4
3 4
3 4
3 4
4
5
=
∑
r =0
5
2
Cr x
3
5− r
7
y
4
r
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5
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2p 3p
iii) −
7
5
6
6
2p
= ∑ (−1) C r
5
r =0
6− r
r 6
3q
7
r
iv)(3 + x – x2)4
81 + 108x − 54x 2 − 96x 3 + 19x 4 + 32x 5 − 6x 6 − 4x 7 + x 8
2
...
6 term in +
2
3
9
i)
2x
6th term in +
2
3
th
9
2x 3y
The general term in + is
2
3
2x
Tr +1 = 9 Cr
3
9− r
3y
2
r
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Put r = 5
4
5
4
5
2x 3y
2 3
T6 = C5 = 9 C5 x 4 y5
3 2
3 x
9
=
9 × 8 × 7 × 6 (24 ) 35 4 5
x y = 189x 4 y5
4
5
1× 2 × 3 × 4 3 2
ii) Ans
...
pq
45
3a
iv) Ans
...
Find the number of terms in the expansion of
3a b
(i) +
4 2
i)
9
3a b
+
4 2
(ii) (3p + 4q)14
(iii) (2x + 3y + z)7
9
Sol
...
3a
b
9
Hence number of terms in + are:
4 2
9 + 1 = 10
iii) (2x + 3y + z)7
Sol
...
2
Hence number of terms in (2x + 3y + z)7 are:
(7 + 1)(7 + 2) 8 × 9
=
= 36
2
2
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4
...
(i) (2 + 3x)–2/3
(ii) (5 + x)3/2
Sol
...
e
...
2
2 2
2
3
i
...
x ∈ − ,
3 3
x
= 5 1 +
5
3/ 2
3/ 2
=5
x
1 +
5
3/ 2
∴ The binomial expansion of (5 + x)3/2 is valid when
x
<1
...
e
...
e
...
sakshieducation
...
sakshieducation
...
Find the (i) 6 term of 1 +
...
Tr+1 in (1 + x)–n = (−1)r
(n)(n + 1)(n + 2)
...
r
Put r = 5, n = 5, x by x/2
T6 = (−1)5
(5)(5 + 1)(5 + 2)(5 + 3)(5 + 4) x
⋅
1⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5
2
5
5
=
−5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 1 5 −63 5
⋅ ⋅ x =
⋅x
1⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 2
16
x2
ii) 7 term of 1 −
3
−4
th
Sol
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(n + r − 1) r
⋅x
1 ⋅ 2 ⋅ 3 ⋅
...
Sol
...
(1)
−2 / 3
th
The general term of (1 – x)
–p/q
(p)(p + q)(p + 2q) +
...
sakshieducation
...
sakshieducation
...
[2 + (9 − 1)3] 4
T10 =
x
1⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9
9
2 ⋅ 5 ⋅ 8 ⋅
...
(26) 4x 9
9!
9
7/4
th
Sol
...
+ [p − (r − 1)q] x
Tr +1 =
(r)!
q
Here p = 7, q = 4, r = 4,
∴ T5 of 1 +
8y
21
x (8y / 21) 2y
=
=
q
4
21
7/4
is
(7)(7 − 4)(7 − 2 × 4)(7 − 3 × 4) 2y
=
1× 2 × 3 × 4
21
=
r
7(3)(−1)(−5) 2 4 y 4
y
⋅
= 70
4
1× 2 × 3 × 4 (21)
21
∴ 5th term of 7 +
8y
∴ T5 in 7 +
3
8y
3
4
4
7/4
7/4
=7
y
21
4
is 77 / 4 (70)
7/4
y
(70)
21
4
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6
...
5
−7 / 3
5
= (3) −7 / 3 1 + x
3
Sol
...
11 q
1⋅ 2 q
Here p = 7, q = 3,
x (5 / 3)x 5
=
= x
q
3
9
∴ (3 + 5x)–7/3 =
(3)
−7 / 3
7 5 (7)(10) 5 2
x +
...
9
81
∴ The first 3 terms of (3 + 5x)–7/3 are
−7 / 3
3
−37 / 3 ⋅ 35x −7 / 3 875 2
,
,3
x
9
81
ii) (1 + 4x)–4 Try your self
iii) (8 – 5x)2/3
5
Sol
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sakshieducation
...
sakshieducation
...
1⋅ 2 q
q
Here X =
5x
X (5x / 8) 5x
, p = 2, q = 3, =
=
8
q
3
24
(1 − X)
∴(8 – 5x)2/3 =
2
5x (2)(2 − 3) 5x
4 1 − 2 +
−
...
12 24
∴ The first 3 terms of (8 – 5x)2/3 are
4,
−5x −25 2
,
x
3 144
iv) (2 – 7x)–3/4 Try your self
7
...
Write (4 + 5x)
2 −3 / 2
= (2 )
–3/2
5
= 4 1 + x
4
−3 / 2
5 −3 / 2 1 5 −3 / 2
1 + x
= 1 + x
4
8 4
General term of (1 + x)–p/q is
Tr+1 = (–1)r
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−1/ 2
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(p)(p + q)(p + 2q) +
...
[3 + (r − 1)2] 5x
(−1) r
8
8
r!
r
3 ⋅ 5 ⋅ 7
...
General term of (1 – x)–n is
Tr +1 =
(n)(n + 1)(n + 2)
...
r
4x
iii) 1 +
5
5/ 2
Sol
...
+ [p − (r − 1)q] x
is Tr +1 =
(r)!
q
−1/ 2
Sol
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sakshieducation
...
sakshieducation
...
+ [p − (r − 1)q] x
is Tr +1 =
(r)!
q
r
8
...
i) Here n = 19 is an odd integer
...
e
...
Hence the largest binomial coefficient is
n
C n i
...
2
24
C12
9
...
Sol
...
There is only one largest binomial coefficient and it is
n
C(n / 2) = 22 C11 = 22 Cr ⇒ r = 11
∴13Cr = 13C11 = 13C2 =
13 ×12
= 78
1× 2
14
4 x2
10
...
2
x
th
Sol
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sakshieducation
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sakshieducation
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Find the 3rd term from the end in the expansion of x −2 / 3 −
8
3
...
Comparing with (X + a)n, we get
X = x −2 / 3 , a =
−3
,n = 8
x2
In the given expansion x −2 / 3 −
8
3
, we have n + 1 = 8 + 1 = 9 terms
...
∴ T7 = n C6 ( X )
n −6
(a 6 )
6
36
−3
= 8C6 (x −2 / 3 )8−6 2 = 8C6 x −4 / 3 ⋅ 12
x
x
=
8 × 7 6 −4 / 3−12
⋅3 ⋅ x
= 28 ⋅ 36 ⋅ x −40 / 3
1× 2
20
1
12
...
x
9
10
1
x
Sol
...
sakshieducation
...
sakshieducation
...
Thus there is no term
containing x9 in the expansion of the given expression
...
Now to find the coefficient of x10
...
13
...
3 2x
10 − r
x
Sol
...
sakshieducation
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sakshieducation
...
Find the set E of x for which the binomial expansions for the following are valid
(i) (3 – 4x)3/4
(ii) (2 + 5x)–1/2
(iii) (7 – 4x)–5
(iv) (4 + 9x)–2/3
(v) (a + bx)r
Sol
...
e
...
3
3
4
−3 3
i
...
E = ,
4 4
ii) (2 + 5x)
−1/ 2
=2
−1/ 2
5x
1 +
2
−1/ 2
The binomial expansion of (2 + 5x)–1/2 is valid when
5x
2
< 1 ⇒| x |<
3
5
−2 2
,
5 5
i
...
E =
iii) (7 − 4x)−5 = 7 −5 1 −
4x
7
−5
The binomial expansion of (7 – 4x)–5 is valid when
4x
7
< 1 ⇒| x |<
7
4
−7 7
i
...
E = ,
4 4
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iv) (4 + 9x)
−2 / 3
=4
−2 / 3
9x
1 +
4
−2 / 3
The binomial expansion of (4 + 9x)–2/3 is valid when
4
9x
< 1 ⇒ |x| <
9
4
−4 4
⇒ x ∈ ,
9 9
−4 4
i
...
E = ,
9 9
v) For any non zero reals a and b, the set of x for which the binomial expansion of (a + bx)r is valid
|a| |a|
,
...
Find the
x
i) 9 term of 2 +
3
−5
th
ii) 10th term of 1 −
3x
4
5x
iii)8 term of 1 −
2
4/5
−3 / 5
th
iv)6th term of 3 +
i)
x
9 term of 2 +
3
2x
3
3/ 2
−5
th
x
Sol
...
(1)
6
−5
with (1 + x)–n,
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we get X = x/6, n = 5
The general term in the binomial expansion of (1 + x)–n is
Tr +1 = (−1)n
(n + r −1)
Cr ⋅ x r
Put r = 8
8
x
T9 = (−1)8 (5+8−1) C8 ⋅ x 8 = 12 C8
6
x
From (1), the 9 term of 2 +
3
−5
th
=2
−5 13
is
8
x 495 x
C8 =
⋅
6 32 6
ii) 10th term of 1 −
3x
4
3x
Sol
...
4
q 20
The general term in (1 – x)p/q is
(−1)4 [ p(p − q)(p − 2q)
...
(4 − 40)] 3x 9
T10 =
9!
20
(−10(−6)(−11)(−16)(−21)
9
(−26)(−31)(−36)
3x
= −4
9!
20
=
−4 × 1× 6 × 11×16 × 21× 26 × 31× 36 3x
9!
20
9
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5x
iii) 8 term of 1 −
2
−3 / 5
th
5x
Sol
...
2
q
5
2
The general term in (1 – x)–p/q is
Tr +1
[ p(p + q)(p + 2q)
...
[3 + (7 − 1)5] x
T8 =
7!
2
=
(3 ⋅ 8 ⋅13 ⋅18 ⋅ 23 ⋅ 28 ⋅ 33) x
7!
2
2x
iv) 6 term of 3 +
3
7
7
3/ 2
th
2x
Sol
...
(1)
3/ 2
with (1 + x)p/q, we get
2x
x (2x / 9) x
, p = 3, q = 2 ⇒ =
=
9
q
2
9
The general term of (1 + x)p/q is
Tr +1
[ p(p − q)(p − 2q)
...
sakshieducation
...
sakshieducation
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Write the first 3 terms in the expansion of
−5
x
(i) 1 + ,
2
i)
x
1 +
2
(ii) (3 + 4x)–2/3 ,
(iii) (4 – 5x)–1/2
−5
Sol
...
1⋅ 2
5
2
5x (5)(6) x
x
∴ 1 + = 1 −
+
−
...
2 4
∴ The first terms in the expansion of
x
1 +
2
−5
are 1,
−5x 15 2
, x
2 4
ii) (3 + 4x)–2/3
Sol
...
sakshieducation
...
sakshieducation
...
1q
1⋅ 2 q
2
2 4x 2 ⋅ 5 4x
= 1− ⋅
+
−
...
9 81
i
...
3−2 / 3 , −3−2 / 3−2 (8x),3−2 / 3−4 (80x 2 )
⇒ 3−2 / 3 − 3−8 / 3 (8x),3−14 / 3 (80x 2 )
iv) (4 – 5x)–1/2
Sol
...
1q
1⋅ 2 q
−p / q
5
4
Here p = 1, q = 2, X = x ⇒
5
∴ 1 − x
4
−1/ 2
= 1+
x 5
= x
q 8
2
1 5x 1 ⋅ 3 5x
= 1+ +
+
...
8 128
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From (1),
5x 75 2
(4 − 5x) −1/ 2 = 2−1/ 2 1 +
+
x +
...
Write the general term of
x
(i) 3 +
2
−2 / 3
(ii) 2 +
(iii) (1 – 4x)–3
x
Sol
...
(1)
The general term of (1 + x)–p/q
Tr +1 = ( −1)
r
( p )( p + q )( p + 2q )
...
sakshieducation
...
sakshieducation
...
[2 + (r − 1)3] x r
r!
18
1 (−1) r (2)(5)(8)
...
2 +
4
r
3x
= 2 1 +
8
3x
= 24 / 5 1 +
8
4/5
4/5
…(1)
Tr+1 of (1 + X)p/q is
Tr +1 =
[ p(p − q)(p − 2q)
...
(4 − (r − 1)5) 3x
Tr +1 =
r!
40
4(−1)(−6)
...
(5r − 9) 3x
r!
40
r
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3x
∴ The general term of 2 +
4
2
4/5
4/5
r −1 4 ⋅1 ⋅ 6
...
(1 − 4x) −3 = (1 − X) − n , here X = 4x, n = 3
...
(2 − 3x)
−1/ 3
3
= 2 1 − x
2
=2
−1/ 3
3
1 − x
2
−1/ 3
−1/ 3
General term of (1 – x)–p/q
( p )( p + q )( p + 2q )
...
sakshieducation
...
sakshieducation
...
[1 + (r − 1)3] x r
r!
2
1 (1)(4)(7)
...
Find the coefficient of
i) x
−6
10
4
in 3x −
x
13
ii) x11 in 2x 2 +
3
x3
2
iii) x in 7x 3 − 2
x
9
2
iv) x
i)
x
−7
−6
2x 2
5
in
− 5
4x
3
7
10
4
in 3x −
x
10
4
Sol
...
(1)
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For coefficient of x–6 , put 10 – 2r = –6
⇒ 2r = 10+6 = 16 ⇒ r = 8
Put r = 8 in (1)
T8+1 = (−1)8 10C8 310−8 (4)8 x10−16 = 10C8 32 48 x −6
10
4
∴ Coefficient of x in 3x −
x
–6
10
is
C8 32 48 = 10 C2 32 48
=
10 × 9
× 9 × 48 = 405 × 48
1× 2
13
11
ii) x
3
in 2x 2 + 3
x
13
Sol
...
(1)
For coefficient of x11, put 26 – 5r = 11
⇒ 5r = 15 ⇒ r = 3
Put r = 3 in (1)
T3+1 = 13C3 (2)10 (3)3 x 26−15
T4 =
13 × 12 × 11 10 3 11
⋅ 2 ⋅3 ⋅ x
1× 2 × 3
∴ Coefficient of x11 in 2x 2 +
13
3
x3
is:
(286)(210)(33)
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9
2
iii) x in 7x 3 − 2 Ans
...
x
7
7
2x 2
5
Sol
...
(1)
x
4
For coefficient of x–7 , put 14 – 7r = –7
⇒ 7r = 21 ⇒ r = 3
Put r = 3 in equation (1)
4
3
4
3
2 5
T3+1 = (−1)3 7 C3 x14−21
3 4
=
−7 × 6 × 5 2 5 −7
x
1× 2 × 3 3 4
7
2x 2
5
∴ Coefficient of x in
− 5 is:
4x
3
–7
= −35 ×
1 53 −4375
⋅ =
324
34 22
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2
...
The general term in
− 2
x
3
10 − r
Tr+1 = (−1)
r 10
x1/ 2
Cr
3
4
2
x
r
5−
r
2
10 − r
1
= (−1)r 10 Cr
3
(4) r ⋅ x
10 − r
= (−1)
r 10
= (−1)
r 10
is
⋅ x −2r
(4)
r
r
5− − 2r
2
⋅x
(4)
1
Cr
3
r
10 −5r
⋅x 2
10 − r
1
Cr
3
...
(1)
8
1
T2+1 = (−1) 2 10 C2 42 ⋅ x 0
3
T3 =
80
729
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3
ii) 3 + 5 x
x
25
3
+5 x
x
25
Sol
...
(1)
For term independent of x, put
−50 + 5r
= 0 ⇒ 5r = 50 ⇒ r = 10
6
Put r = 10 in equation (1),
T10+1 = 25C10 (3)15 (5)10 x 0
T11 = 25C10 (3)15 (5)10
i
...
14
iii) 4x 3 +
7
x2
14
7
Sol
...
(1)
www
...
com
www
...
com
For term independent of x,
Put 4x – 5r = 0 ⇒ r = 42/5 which is not an integer
...
2x 2 15
iv)
+
5
4x
9
Ans
...
Find the middle term(s) in the expansion of
10
3x
(i) − 2y
7
(iii) (4x + 5x )
2
11
3
(ii) 4a + b
2
3 17
3
(iv) 3 + 5a 4
a
20
Sol
...
e
...
2
2
10
i)
3x
− 2y
7
Sol
...
i
...
10
th
+ 1 = 6 term
2
www
...
com
www
...
com
10
3x
∴ T6 in − 2y
7
is :
5
35 5
3x
5
10
= C5 (−2y) = −( C5 ) 5 ⋅ 2 (xy)5
7
7
10
5
6
= − C5 x 5 y 5
7
10
11
3
ii) 4a + b
2
Sol
...
e
...
11
3
T6 in 4a + b
2
is:
5
3
35
= C5 (4a) b = 11C5 (4)6 5 a 6 b5
2
2
11
=
6
11× 10 × 9 × 8 × 7 7 5 6 5
2 ⋅3 ⋅a b
1× 2 × 3 × 4 × 5
= 77 × 28 × 36 × a 6 b5
11
3
T7 in 4a + b is:
2
6
36
3
= 11C6 (4a)5 b = 11C5 (4)5 6 a 5 b6
2
2
=
11× 10 × 9 × 8 × 7 4 6 5 6
2 ⋅3 ⋅a b
1× 2 × 3 × 4 × 5
= 77 × 25 × 37 × a 5 b 6
www
...
com
n +1
n +3
and
terms = 7th and 7th
2
2
www
...
com
iii) (4x 2 + 5x 3 )17 Try yourself
...
Fin the numerically greatest term (s) in the expansion of
i) (4 + 3x)15 when x =
7
2
ii)(3x + 5y)12 when x =
iv) (3 + 7x)n when x =
iii)(4a – 6b)13 when a = 3, b = 5
i)
(4 + 3x)15 when x =
1
4
and y =
2
3
4
, n = 15
5
7
2
15
3
Sol
...
sakshieducation
...
sakshieducation
...
Write (3x + 5y) = 3x 1 +
3x
12
12
12 12
5 y
= 3 x 1 +
3x
On comparing 1 +
n = 17, x =
12
5 y
n
with (1 + x) , we get
3x
5 y 5 (4 / 3) 5 8 40
⋅ =
= ⋅ =
3 x 3 (1/ 2) 3 3 9
40
(12 + 1)
(n + 1) | x |
9
=
Now
40
1+ | x |
1+
9
=
13 × 40 520
30
=
= 10
49
49
49
Which is not an integer
...
sakshieducation
...
sakshieducation
...
G
...
G
...
Write (4a – 6b) = 4a 1 −
4a
13
13
3 b
= (4a)13 1 −
2a
13
3 b
On comparing 1 −
2a
We get n = 13, x =
x=
with (1 + x)n
−3 b
2 a
−3 5 −5
× =
2 3 2
www
...
com
www
...
com
Now
=
(n + 1) | x |
=
1+ | x |
−5
5
14 ×
2
2
=
5
−5
1+
1+
2
2
(13 + 1)
70
= 10 which is an integer
...
13
9
3 b
3 b
13
T10 in 1 −
= C9 − ⋅
2 a
2a
9
3 5
5
= C9 ⋅ = 13C9
2 3
2
9
13
T10 in (4a – 6b)13 is
9
5
5
= (4a)13 ⋅ 13C9 = (4 × 3)13 ⋅ 13C9
2
2
9
9
5
= C9 (12) (12) = 13C9 (12)4 (30)9
2
13
4
9
13
3 b
T11 in 1 −
2a
10
−3 b
is = C10 ⋅
2 a
13
10
10
3 5
5
= 13C10 × = 13C10
2 3
2
∴ N
...
term in (4a – 6b)13 is
10
10
5
5
= (4a)13 ⋅ 13C10 = (4 × 3)13 ⋅ 13C10
2
2
= (12)13 ⋅ 13C10
510 13
510
= C10 (12)3 ⋅ (12)10 ⋅ 10
210
2
= 13C10 (12)3 (30)10
www
...
com
www
...
com
iv) (3 + 7x)n when x =
4
, n = 15 Try your self
5
5
...
+ (3n + 2) ⋅ Cn
= (3n + 4) ⋅ 2n −1
...
Let S = 2 ⋅ C0 + 5 ⋅ C1 + 8 ⋅ C2 +
...
S = (3n + 2)C0 + (3n − 1)C1 + (3n − 4)C2 +
...
+ (3n + 4)C n
Adding = (3n + 4)(C0 + C1 + C 2 +
...
= 0
Sol
...
P
...
(n + 1)terms
= C0 − 4 ⋅ C1 + 7 ⋅ C 2 − 10 ⋅ C3 +
...
= 0
www
...
com
www
...
com
C1 C3 C5 C7
2n − 1
iii)
+
+
+
+
...
2
4
6
8
Sol
...
2
4
6
8
=
n n(n − 1)(n − 2) n(n − 1)(n − 2)(n − 3)(n − 4)
+
+
+
...
+
...
n +1
1 (n +1)
1
2n − 1
(n +1)
(n +1)
(n +1)
=
2n − 1 =
=
C0 +
C2 +
C4 +
...
=
2
4
6
8
n +1
3
2
9
3
iv) C0 + C1 + C2 +
27
3n
C3 +
...
Let S =
3
32
33
3n
C0 + C1 + C2 + C3 +
...
+ Cn
...
+ (n + 1)C n ⋅
2
3
3
n +1
www
...
com
www
...
com
⇒ (n + 1)3 ⋅ S
= (n +1) C1 ⋅ 3 + (n +1) C2 ⋅ 32 + (n +1) C3 ⋅ 33 +
...
+ 2n ⋅ Cn = 3n
Sol
...
H
...
= C0 + 2 ⋅ C1 + 4 ⋅ C2 + 8 ⋅ C3 +
...
+ Cn (2n )
= (1 + 2)n = 3n
[ (1 + x) n = C0 + C1 ⋅ x + C2 x 2 +
...
Using binomial theorem, prove that 50n – 49n – 1 is divisible by 492 for all positive integers
n
...
50n – 49n – 1 = (49 + 1)n – 49n – 1
= [ n C0 (49)n + n C1 (49)n −1 + n C2 (49)n −2 +
...
+ n Cn −2 (49)2 + (n)(49) + 1 − 49n − 1
= 492 [(49)n −2 + n C1 (49)n −3 + n C 2 (49) n −4 +
...
www
...
com
www
...
com
7
...
Sol
...
+ 2n C2n − 2 (26) 2 − 2n C2n −1 (26) + 2n C2n (1)] + 52n − 1
= 2n C0 (26)2n − 2n C1 (26) 2n −1 + 2n C2 (26) 2n −2 −
...
+ 2n C2n −2 ]
is divisible by (26)2 = 676
∴54n + 52n – 1 is divisible by 676, for all positive integers n
...
If (1 + x + x 2 ) n = a 0 + a1x + a 2 x 2 +
...
+ a 2n = 3n
ii) a 0 + a 2 + a 4 +
...
+ a 2n −1 =
3n − 1
2
iv) a 0 + a 3 + a 6 + a 9 +
...
(1 + x + x 2 ) n = a 0 + a1x + a 2 x 2 +
...
sakshieducation
...
sakshieducation
...
+ a 2n = 3n
ii)(1) + (2) ⇒2( a 0 + a 2 + a 4 +
...
+ a 2n =
3n + 1
2
iii) (1) – (2) ⇒ 2(a1 + a 3 + a 5 +
...
+ a 2n −1 =
3n − 1
2
iv) Put x = 1
a 0 + a1 + a 2 +
...
+ a 2n ω2n = 0 …(b)
Put x = ω2
a 0 + a1ω2 + a 2ω4 + a 3ω6 +
...
+ a 2n (1 + ω2n + ω4n ) = 3n
⇒ 3a 0 + a1 (0) + a 2 (0) + 3a 3 +
...
=
3n
= 3n −1
3
www
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com
www
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com
9
...
Sol
...
(2)
⇒ Coefficients are equal
⇒ 21C2r+3 = 21C3r+3
⇒ 21 = (2r + 3) + (3r + 3) (or) 2r + 3 = 3r + 3
⇒ 5r = 15 ⇒ r = 3 (or) r = 0
Hence r = 0, 3
...
If the coefficients of x
10
1
in the expansion of ax 2 +
bx
is equal to the coefficient of
11
x
–10
1
in the expansion of ax − 2 ; find the relation between a and b where a and b are real
bx
numbers
...
The general term in the expansion of ax 2 +
bx
1
Tr +1 = 11Cr (ax 2 )11−r
bx
= Cr a
11
11− b
is
r
r
1 22−2r −r
x
b
To find the coefficient of x10, put
22 – 3r = 10 ⇒ 3r = 12 ⇒ r = 4
11
1
Hence the coefficient of x in ax 2 +
bx
10
4
a7
1
is = C7 ⋅ a = 11C7 4
b
b
11
7
www
...
com
...
sakshieducation
...
Hence from (1) and (2), we get
11
C7 ⋅
a7
a4
= − 11C7 ⋅ 7
a4
b
⇒ a3 =
−1
⇒ a 3b3 = −1 ⇒ ab = −1
3
b
20
11
...
The general term in the expansion of x 2 −
2x
2 20 − r
Tr +1 = Cr (x )
20
−1
2x
1
, find Tk and Tk+3
...
e
...
www
...
com
www
...
com
∴ k = 11
Put r = 10 in eq
...
If the coefficients of (2r + 4)th and (r – 2)nd terms in the expansion of (1 + x)18 are equal,
find r
...
T2r+4 term of (1 + x)18 is
T2r + 4 = 18C 2r +3 (x)2r +3
Tr −2 term of (1 + x)18
Tr −2 = 18Cr −3 (x) r −3
Given that the coefficients of (2r + 4)th term = The coefficient of (r – 2)nd term
...
Sol
...
(1 − 2x) 2
1 + 2x
= (1 + 2x)(1 − 2x) −2
2
(1 − 2x)
= (1 + 2x)[1 + 2(2x) + 3(2x)2 + 4(2x)3 + 5(2x)4 + 6(2x)5 + 7(2x)6 + 8(2x)7 + 9(2x)8 + 10(2x)9
+11(2x)10 +
...
]
∴ The coefficient of x10 in
1 + 2x
is
(1 − 2x) 2
= (11)(2)10 + 10(2)(29 ) = 210 (11 + 10) = 2 × 110
www
...
com
www
...
com
14
...
Sol
...
+ [p − (r − 1)q] x
Tr +1 =
(r)!
q
Here p = 3, q = 5,
r
X 4x
=
q 5
Put r = 4
T4+1 =
(3)(3 + 5)(3 + 2 × 5)(3 + 3 × 5) 4x
1× 2 × 3 × 4
5
4
∴ Coefficient of x4 in (1 – 4x)–3/5 is
4
(3)(8)(13)(18) 4
234 × 256 59904
=
=
1× 2 × 3 × 4 5
625
625
15
...
3 3⋅ 6 3⋅ 6 ⋅9
Sol
...
1 3 1⋅ 2 3 1⋅ 2 ⋅ 3 3
The series of the right is of the form
2
3
p x p(p + q) x p(p + q)(p + 2q) x
1+ +
+
+
...
sakshieducation
...
sakshieducation
...
4 4 ⋅ 8 4 ⋅ 8 ⋅12
Sol
...
4 4 ⋅ 8 4 ⋅ 8 ⋅12
2
3
3 1 3⋅5 1 3⋅5⋅ 7 1
= ⋅ +
+
+
...
1 4 1⋅ 2 4
Comparing (1 + S) with
2
(1 − x)
−p / q
p x p(p + q) x
= 1+ +
+
...
5 ⋅10 5 ⋅10 ⋅15
4
5
Sol
...
5 ⋅10 5 ⋅10 ⋅15
2
3
4 1 4 ⋅ 7 1 4 ⋅ 7 ⋅10 1
= 1+ − +
− +
− +
...
1q
1⋅ 2 q
www
...
com
www
...
com
Here p = 4, q = 3,
∴ S = (1 − x)
5
=
8
4/3
=
−p / q
x
1
−3
=− ⇒x=
q
5
5
3
= 1 +
5
−4 / 3
8
=
5
−4 / 3
54 / 3 3 54 3 625
= 4 =
16
84 / 3
2
3
4 4 ⋅ 7 4 ⋅ 7 ⋅10
625 54 / 3
∴1 − +
−
+
...
4 ⋅ 8 4 ⋅ 8 ⋅12 4 ⋅ 8 ⋅12 ⋅16
Sol
...
4 ⋅ 8 4 ⋅ 8 ⋅12 4 ⋅ 8 ⋅12 ⋅16
1 ⋅ 3 1⋅ 3 ⋅ 5
1⋅ 3 ⋅ 5 ⋅ 7
−
+
−
...
4
4 4 ⋅ 8 4 ⋅ 8 ⋅12
2
3
3
1 1 1⋅ 3 1 1⋅ 3 ⋅ 5 1
⇒ + S = 1− ⋅ +
−
+
...
1q
1⋅ 2 q
1⋅ 2 ⋅ 3
q
1
= 1 +
2
= (1 + x)
−p / q
∴ S=
−1/ 2
3
=
2
−1/ 2
2
3
2 3
−
3 4
=
www
...
com
x 1
1
= ⇒x=
q 4
2
www
...
com
16
...
(i)
5
(ii) 7 127
242
(iii)
5
32
...
i)
5
242 = (243 − 1)
1/ 5
1 1
= (35 )1/5 1 − ⋅
+
5 243
= (243)
1/5
1
⋅ 1 −
243
11
− 1 1 2
55
−
...
00243) − (0
...
25
5
5
1
1
∵
= = (0
...
00243
243 3
3
6
≃ 3 − (0
...
00243)2 −
...
001458 − 0
...
998541
ii)
7
127
iii) 5 32
...
If |x| is so small that x2 and higher powers of x may be neglected then find the approximate
values of the following
...
=
2
(3 − 2x)2
2
3 1 − 3 x
www
...
com
www
...
com
1/ 2
2 3
= 1 + x
9 4
2
1 −
3
−2
2 1 3
2
= 1 + ⋅ x 1 − (−2) x
9 2 4
3
(After neglecting x2 and higher powers of x)
2 3 4 2 3
4
= 1 + x 1 + x = 1 + x + x
9 8 3 9 8
3
(Again by neglecting x2 term)
2 41 2 41
= 1 + x = +
x
9 24 9 108
∴
(4 + 3x)1/ 2 2 82
2 41
= +
x= +
x
2
9 108
9 108
(3 − 2x)
3/ 2
ii)
2x
1/ 5
1 −
(32 + 5x)
3
(3 − x)3
3/ 2
2x
1/ 5
1 −
(32 + 5x)
3
Sol
...
sakshieducation
...
sakshieducation
...
Suppose s and t are positive and t is very small when compared to s, then find an
1/ 3
s
approximate value of
s+t
1/ 3
s
−
s−t
...
Since t is very small when compared with s, t/s is very small
...
1⋅ 2
3!
3 s
s
s
1 1
1 1 1
1 t − 3 − 3 − 1 t 2 − 3 − 3 − 1 − 3 − 2 t 3
= 1 − − +
−
+
...
sakshieducation
...
sakshieducation
...
Suppose p, q are positive and p is very small when compared to q
...
Do it yourself
...
20
...
3 x 2 + 64 − 3 x 2 + 27
= (64 + x 2 )1/ 3 − (27 + x 2 )1/ 3
1/ 3
1/ 3
x2
1 +
64
= (64)
1/ 3
1/ 3
x2
1 +
27
− (27)
x2 x2
= 4 1 +
− 3 1 +
192 81
(By neglecting x4 and higher powers of x)
= 4+
x2
x2
(27 − 48) 2
−3−
= 1+
x
48
27
48 × 27
7x 2
7 2
−21 2
= 1+
= 1−
x
x = 1−
432
432
48 × 27
∴ 3 x 2 + 64 − 3 x 2 + 27 = 1 −
7 2
x
432
21
...
Sol
...
+ r − 1 2 r
22 2
2 2 2
+
...
+
r
1 3
1⋅ 2 3
(1 ⋅ 2 ⋅ 3
...
(2r + 1) 2
= 1+
+
...
2
1 ⋅ 2 3 (1 ⋅ 2)2 3
(1 ⋅ 2 ⋅
...
sakshieducation
...
www
...
com
2
r
3 ⋅ 5 ⋅ 7
...
+
+
...
Prove that 2 ⋅ C0 + 7 ⋅ C1 + 12 ⋅ C 2 +
...
First method:
The coefficients of C0, C1, C2, …… Cn are in A
...
with first term a = 2, C
...
(d) = 5
∵ a ⋅ C0 + (a + d)C1 + (a + 2d)C2 +
...
H
...
i
...
Tr +1 = (5r + 2)Cn
23
...
+ 3n Cn = 4n
ii)
C1
C
C
C
n(n + 1)
+ 2 ⋅ 2 + 3 ⋅ 3 +
...
(i) We have
(1 + x) n = C0 + C1x + C2 x 2 +
...
+ Cn 3n
∴ C0 + 3C1 + 32 C 2 +
...
sakshieducation
...
sakshieducation
...
+ n n
C0
C1
C2
Cn −1
n C 2 n C3
n Cn
C1
=n
+ 2 n + 3 n +
...
+ n
1
2
3
n
= n + (n − 1) + (n − 2) +
...
+ n =
n(n + 1)
2
24
...
+ C n −r ⋅ Cn = 2n Cn + r and hence
deduce that
i)
2
2
2
2
C0 + C1 + C2 +
...
+ Cn −1Cn = 2n Cn +1
Sol
...
+ Cn x n …(1)
On replacing x by 1/x in the above equation,
n
C1 C 2
Cn
1
+ 2 +
...
(2)
1 + = C0 +
x x
x
x
From (1) and (2)
n
C1 C2
Cn
1
n
1 + (1 + x) = C0 + + 2 +
...
+ Cn x n )
...
H
...
of (3)
= C0Cr + C1C r +1 + C2Cr + 2 +
...
sakshieducation
...
sakshieducation
...
H
...
of (3)
= the coefficient of xr in
(1 + x)2n
xn
= the coefficient of xn+r is (1 + x)2n
= 2nCn+r
From (3) and (4), we get
C0 ⋅ C1 + C1 ⋅ C 2 + C2 ⋅ C3 +
...
+ Cn = 2n C n
ii) On substituting r = 1 in (i) we get
C0 ⋅ C1 + C1 ⋅ C 2 + C2 ⋅ C3 +
...
+ (4n + 3)C2
2
n
= (2n + 3) 2n Cn
2
2
2
Sol
...
+ (4n − 1)C 2 −1 + (4n + 3)Cn
...
, on writing the terms of R
...
S
...
+ 7C2 −1 + 3Cn ……(2)
n
Add (1) and (2)
2
2
2
2S = (4n + 6)C0 + (4n + 6)C1 +
...
+ C2 )
2
n
= 2(2n + 3) 2n Cn
∴ S = (2n + 3) 2n Cn
www
...
com
www
...
com
25
...
Write (2 + 3x)10 = 2 1 + x = 210 1 +
2
2
10
3x
First find N
...
term in 1 +
2
3x
Let X =
=
2
11
8 = 33
2
16
3×
Now consider
33
(10 + 1)
(n + 1) | x |
16 = 11× 33 = 363
=
33
1+ | x |
48
48
+1
16
363
Its integral part m =
=7
48
∴ Tm+1 is the numerically greatest term in
10
3x
1 +
2
3x
i
...
T7 +1 = T8 = 10 C7
2
7
7
3 11
33
= C7 × = 10 C7
2 8
16
7
10
7
33
...
G
...
sakshieducation
...
sakshieducation
...
14
4y
Sol
...
Here | T5 | = | T6 | are N
...
terms
...
G
...
They are
4
1
T5 = C 4 (24)14
2
14
5
1
T6 = − C5 (24)14
2
14
But | T5 | = | T6 |
www
...
com
www
...
com
26
...
Sol
...
+ n C n −2 (35)2 + n Cn −1 (35)1 + n Cn − 35n − 1
= (35)n + n C1 (35) n −1 + n C2 (35)n −2 +
...
+ n Cn −2
= 1225 (k), for same integer k
...
27
...
Sol:
We know that
(4x − 7y)49 = 40C0 (4x)49 − 49C1 (4x) 48 (7y) + 49C2 (4x)47 (7y)2 − 49C3 (4x) 46 (7y)3 +
...
(1)
(4x + 7y)49 = 40C0 (4x) 49 + 49C1 (4x)48 (7y) + 49C 2 (4x) 47 (7y)2 + 49C3 (4x)46 (7y)3 +
...
(2)
(1) + (2) ⇒
(4x − 7y) 49 + (4x + 7y)49 = 2[ 49 C0 (4x) 49 + 49C 2 (4x) 47 (7y)2 + 49C4 (4x)45 (7y) 4 +
...
28
...
Sol:
The last 20 coefficients in the expansion of (1 − x)39 are
30
C 20 , 39C 21,
...
We know that
∴ 39 C0 + 39 C1 + 39 C2 +
...
+ 39 C39 = 239
⇒ 39 C39 + 39 C38 + 39 C37 +
...
+ 39 C39 = 239
(∵ n Cr = n Cn −r )
⇒ 2[39 C20 + 39C 21 + 39C22 +
...
+ 39C39 ] = 238
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∴ The sum of last 20 coefficients in expansion of (1 + x)39 is 238
...
If A and B are coefficients of xn in the expansion of (1 + x)2n and (1 + x)2n–1 respectively,
then find the value of A/B
...
Sol:
Coefficient of xn in the expansion of
(1 + x)2n −1 is
2n −1
Cn
...
B
30
...
15 3 +
...
15 15
15
C0
C1
C2
C14
15
i)
ii) C0
...
C4 + C 2
...
+ Cn −3
...
+ (n + 2)2 Cn
iii)
iv) 3C0 + 6C1 + 12C2 +
...
2 n Cn
Sol:
i)We know that
Cr
n!
(r − 1)!(n − r + 1)!
=
×
n!
Cr −1 (n − r)!r!
n
n
=
n − r +1
r
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15
15
15
15
C3
C
C1
C2
∴ 15 + 2 15 + 3
...
+ 15
...
+ 10 ×
1
10
2 3
= 15 + 14 + 13 +
...
+ Cn x n …(1)
n
( x + 1)n = C0 x n + C1x n −1 + C2 x n −2 +
...
+ Cn x n )
(C0 x n + C1x n −1 + C2 x n −2 +
...
+ C n −3 ⋅ C n
i
...
C0 ⋅ C3 + C1 ⋅ C4 + C2 ⋅ C5 +
...
+ (n + 2)2 Cn
n
= ∑ ( r + 2 ) Cn
2
r =0
n
(
)
= ∑ r 2 + 4r + 4 Cr
r =0
n
n
n
r =0
r =0
r =0
= ∑ r 2 Cr + 4 ∑ rC r + 4 ∑ Cr
n
n
n
n
r =0
r =0
r =0
r =0
n
n
n
r =2
r =1
r =0
= ∑ r ( r − 1) Cr + ∑ rC r + 4 ∑ r Cr + 4 ∑ Cr
= ∑ r ( r − 1) Cr + 5∑ rCr + 4 ∑ Cr
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= n ( n − 1) 2n −2 + 5n
...
2n
(
= (n
)
+ 9n + 16 ) 2
= n 2 + 9n + 44 2n −2
2
n −2
...
+ 3
...
+ Cn 2n )
= 3[1 + 2]n
= 3 ⋅ 3n
= 3n +1
...
If 1+ x + x 2 + x 3
)
7
= b0 + b1x + b 2 x 2 + … +b 21x 21 , then find the value of
i) b0 + b2 + b 4 +
...
+ b 21
Sol:
Given
(1+ x + x
2
+ x3
)
7
= b0 + b1x + b 2 x 2 + … +b 21x 21 …(1)
Substituting x = 1 in (1),
We get
b0 + b1 + b 2 +
...
+ b 20 − b 21 = 0
…(3)
i) (2) + (3)
⇒ 2b0 + 2b2 + 2b4 +
...
sakshieducation
...
sakshieducation
...
+ 2b 21 = 47
⇒ b1 + b3 + b5 +
...
32
...
33
...
Sol:
We have 22013
= 2(22012 )
= 2(24 )503
= 2(16)503
= 2(17 − 1)503
= 2[503 C017503 − 503C117502 + 503C217501 −
...
sakshieducation
...
3
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= 2[503 C017503 − 503C117502 + 503C 217501 −
...
∴ 22013 = 17m − 2 (or) 17k + 15
∴ The remainder is –2 or 15
...
If 36, 84, 126 are three successive binomial coefficients in the expansion of (1 + x)n, find n
...
Let
Cr–1,
n
Cr ,
n
Cr+1 are three successive binomial coefficients in the expansion of
(1 + x)n, find n
...
(1)
10
n
⇒
Cr +1 126
n−r 3
=
⇒
=
84
r +1 2
Cr
n
⇒ 2n − 2r = 3r + 3 ⇒ 2n = 5r + 3
...
sakshieducation
...
sakshieducation
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If the 2nd, 3rd and 4th terms in the expansion of (a + x)n are respectively 240, 720, 1080, find
a, x, n
...
T2 = 240 ⇒ nC1 an–1 x = 240 …(1)
T3 = 720 ⇒ nC2 an–2 x2 = 720
...
(3)
n
(2)
C a n −2 x 2 720
⇒ n 2 n −1 =
(1)
240
C1a x
⇒
n −1 x
= 3 ⇒ (n − 1)x = 6a
...
(5)
n −2 2
(2)
720
3 a 2
C2 a x
(4)
(n − 1)x
6a
n −1 2
⇒
=
⇒
=
(5)
2(n − 2)x 9a
2n − 4 3
⇒ 3n − 3 = 4n − 8 ⇒ n = 5
From (4), (5 – 1)x = 6a ⇒ 4x = 6a
⇒x=
3
a
2
Substitute x =
5
3
a , n = 5 in (1)
2
3
3
C1 ⋅ a 4 ⋅ a = 240 ⇒ 5 × a 5 = 240
2
2
a5 =
480
= 32 = 25
15
∴ a = 2, x =
3
3
a = (2) = 3 ∴ a = 2, x = 3, n = 5
2
2
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3
...
P
...
Sol
...
P
...
Find the sum of the coefficients of x and x
3 14− r
Tr +1 = Cr (2x )
3
− 2
x
3
in the expansion of 2x 3 − 2
...
The general term in 2x 3 −
14
–18
3
x2
is:
r
= (−1) r 14 C r (2)14−r ⋅ (3)r ⋅ x 42−r ⋅ x −2r
= (−1) r ⋅ 14 Cr 214−r (3)r x 42−5r
...
sakshieducation
...
sakshieducation
...
5
...
(x + a) n = n C0 x n + n C1x n −1a + n C2 x n − 2a 2 + n C3 x n −3a 3 +
...
) + (n C1x n −1a + n C3 x n −3a 3 + n C5 x n −5a 5 +
...
+ n Cn (−1)n a n
= (n C0 x n + n C 2 x n −2 a 2 + n C4 x n −4 a 4 +
...
)
= P−Q
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i) P2 – Q2 = (P + Q)(P − Q)
= (x + a)n (x – a)n
= [(x + a) (x – a)]n = (x2 – a2)n
ii) 4PQ = (P + Q)2 – (P – Q)2
= [(x + a)n]2 – [(x – a)n]2
= (x + a)2n – (x – a)2n
6
...
Given a1, a2, a3, a4 are the coefficients of 4 consecutive terms in (1 + x)n respectively
...
H
...
H
...
H
...
sakshieducation
...
sakshieducation
...
Prove that (2n C0 ) 2 − (2n C1 )2 + (2n C2 ) 2 − (2n C3 ) 2 +
...
(x + 1)2n = 2n C0 x 2n + 2n C1x 2n −1 +
2n
C2 x 2n −2 +
...
+ 2n C 2n x 2n
2n
Cn
...
(2)
Multiplying eq
...
+ 2n C 2n )
(2n C0 − 2n C1x + 2n C2 x 2 +
...
+ ( 2n C2n )2 = (−1) n
8
...
(Cn −1 + Cn ) =
Sol
...
Cn
n!
(C0 + C1 )(C1 + C2 )(C2 + C3 )
...
Cn−1 1+ n
Cn−1
C0 C1
n C1 n C2 n Cn
= 1+ n 1+ n
...
Cn−1
C0
C1
Cn−1
n n −1 1
= 1+ 1+
...
Cn−1[C0 = Cn ]
1
2 n
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1 + n 1 + n
1+ n
=
...
Cn −1 ⋅ C n
1 2
n
=
(1 + n) n
C1C2
...
(Cn −1 + Cn ) =
⋅ C0 ⋅ C1 ⋅ C2 ⋅
...
Find the term independent of x in (1 + 3x) 1 +
...
(1 + 3x) 1 + = (1 + 3x) n
3x
3x
n
n
n
1
1
= (1 + 3x)2n = n n
3 ⋅x
3x
2n
∑ (2n Cr )(3x)r
r =0
The term independent of x in
n
1
1
(1 + 3x) 1 + is n ( 2n Cn )3n = 2n C n
3
3x
n
10
...
+a 20 x 20 then prove that
i) a 0 + a1 + a 2 +
...
+ a 20 = 410
Sol
...
+a 20 x 20
i)
Put x = 1
(1 + 3 − 2)10 = a 0 + a1 + a 2 +
...
+ a 20 = 210
ii) Put x = –1
(1 − 3 − 2)10 = a 0 − a1 + a 2 +
...
+ a 20 = (−4)10 = 410
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11
...
Sol
...
+ n C n (11)n
C 2 (5 5) n −2 (11)2 +
...
= 2k where k is an integer
...
⇒ F – f is an integer since R is an integer
...
ii) (R + F)F = (R + F)f,
∵F = f
= (5 5 + 11)n (5 5 − 11) n
n
= (5 5 + 11)(5 5 − 11) = (125 − 121)n = 4n
∴ (R + F)F = 4n
...
If I, n are positive integers, 0 < f < 1 and if (7 + 4 3) n = I + f , then show that
(i) I is an odd integer and (ii) (I + f)(I – f) = 1
...
Given I, n are positive integers and
(7 + 4 3) n = I + f , 0 < f < 1
Let 7 − 4 3 = F
Now 6 < 4 3 < 7 ⇒ −6 > −4 3 > −7
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⇒ 1 > 7 − 4 3 > 0 ⇒ 0 < ( 7 − 4 3 )n < 1
∴0
n C0 7 n + n C1 7 n −1 (4 3) + n C2 7 n −2 (4 3) 2
=
+
...
+ n C n (−4 3) n
= 2 n C0 7 n + n C2 7 n −2 (4 3) 2
...
∴ 1 + f + F is an even integer
...
But 0 < f < 1 and 0 < F < 1 ⇒ f + F < 2
∴f+F=1
…(1)
⇒ I + 1 is an even integer
...
(I + f)(I – f) = (I + f)F, by (1)
= (7 + 4 3)n (7 − 4 3) n
n
= (7 + 4 3)(7 − 4 3) = (49 − 48)n = 1
3
3
2
C
(n)(n + 1) 2 (n + 2)
13
...
12
r =1
C r −1
n
n
2
2
n
C
n − r +1
Sol
...
sakshieducation
...
sakshieducation
...
If x =
1⋅ 3 1⋅ 3 ⋅ 5 1⋅ 3 ⋅ 5 ⋅ 7
2
+
+
+
...
3 ⋅ 6 3 ⋅ 6 ⋅ 9 3 ⋅ 6 ⋅ 9 ⋅12
1⋅ 3 1⋅ 3 ⋅ 5 1⋅ 3 ⋅ 5 ⋅ 7
+
+
+
...
Given x =
2
2
1⋅ 3 1 1⋅ 3 ⋅ 5 1
=
+
+
...
− 1 +
1 3 1⋅ 2 3 1 ⋅ 2 ⋅ 3 3
3
Here p = 1, q = 2,
x 1
2
= ⇒x=
q 3
3
= (1 − x)− p / q −
1
=
3
−1/ 2
−
4 2
= 1 −
3 3
−1/ 2
−
4
3
4
4
= 3−
3
3
⇒ 3x + 4 = 3 3
Squaring on both sides
(3x + 4) 2 = (3 3) 2 ⇒ 9x 2 + 24x + 16 = 27
⇒ 9x 2 + 24x = 11
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(1 − 3x) 2
15
...
(3 − x)3 / 2
5
Sol
...
1⋅ 2 ⋅ 3 3
1⋅ 2 ⋅ 3 ⋅ 4 3
2 3 1⋅ 2 3
35x 3
35
77
x 5
(1 − 6x + 9x 2 ) 1 + + x 2 +
+
x4
x 5 +
...
(1 + x)2
2
= (1 + x)2 1 − x
3
3
2
1 − x
3
(1 + x)2
2
1 − x
3
3
...
1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6 3 1⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6 ⋅ 7 3
1⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6 ⋅ 7 ⋅ 8 3
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∴ Coefficient of x in
(1 + x)2
8
8
2
1 − x
3
7
2
2
2
= 45 + 2 × 36 + 28
3
3
3
3
is
6
6
4
2
2
= 45 × + 72 × + 28
9
3
3
6
96 × 26 2048
2
= (20 + 48 + 28) =
=
243
36
3
(2 + 3x)3
iii) Find the coefficient of x in
...
(2 + 3x)3
= (2 + 3x)3 (1 − 3x) −4
4
(1 − 3x)
= (8 + 36x + 54x 2 + 27x 3 )
[1 + 4 C1 (3x) + 5C2 (3x)2 + 6 C3 (3x)3 + 7 C4 (3x)4 + 8C5 (3x)5 + 9 C6 (3x)6 +
...
sakshieducation
...
sakshieducation
...
Find the coefficient of x in the expansion of
...
(1 − 5x)3 (1 + 3x 2 )3 / 2 (1 − 5x)3 (1 + 3x 2 )3 / 2
=
1/ 3
(3 + 4x)1/ 3
4
3 1 + 3
=
=
4
(1 − 5x)3 (1 + 3x 2 )3 / 2 1 +
1/ 3
3
3
1
1
1/ 3
3
−1/ 3
[1 − 15x + 75x 2 − 125x 3 ]
3 3
− 1
3
2 2
1 + (3x 2 ) +
(3x 2 )2 +
...
1⋅ 2
1⋅ 2 ⋅ 3
3
3
3 3
=
1
1/ 3
3
9
(1 − 15x + 75x 2 − 125x 3 ) 1 + x 2 +
...
+ x −
1 −
9 81
2187
=
1
9 2 135 3
2
3
1 − 15x + 75x − 125x + 2 x − 2 x +
...
=
1
159 2 385 3
1 − 15x + 2 x − 2 x +
...
(1 − 5x)3 (1 + 3x 2 )3 / 2
1
∴ Coefficient of x in
is = 1/3
1/ 3
(3 + 4x)
3
3
32 896
385 159 4
− 2 − 2 × 9 − 15 × 81 − 2187
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=
1 385 77274 + 12960 + 896
−
−
2187
3 2
=
1 −841995 − 182260
1024255
= − 3 3(4274)
4374
3
1/3
1/3
17
...
x =
5
5⋅7
5⋅7 ⋅9
2
+
+
+
...
2
3
(2!) ⋅ 3 (3!) ⋅ 3 (4!)3
5
5⋅7
5⋅7 ⋅9
+
+
+
...
2!32 3!33
4!34
=
3⋅ 5 1 3⋅ 5⋅ 7 1 3⋅5⋅ 7⋅ 9 1
3 1
+
+
...
1 3 2! 3
3! 3
2
3 1 3⋅5 1
⇒ 2 + x = 1+ +
+
...
1q
1⋅ 2 q
Here p = 3, q = 2,
−p/q
∴x + 2 = (1− y)
x 2 + 4x + 4 = 27
y 1
q 2
= ⇒y= =
q 3
3 3
−3/2
2
= 1−
3
−3/2
1
=
3
= (3)3/2 = 27 Squaring on both sides
⇒ x 2 + 4x = 23
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18
...
1 +
7
1 1⋅ 3 1 1⋅ 3 ⋅ 5 1
+
+
...
102 1 ⋅ 2 104 1 ⋅ 2 ⋅ 3 106
1 1 1⋅ 3 1
2
1⋅ 3 ⋅ 5 1
3
=1 +
+
+
+
...
02
q 100
100 100
∴1+
1 1⋅ 3 1
+ ⋅ 4 +
...
02)
−1/2
= (0
...
4
6
5 10 1 ⋅ 2 10 1 ⋅ 2 ⋅ 3 10
=
75 2
= 2
5 7
19
...
2
2⋅4
2⋅4⋅6
= 1+
x x(x + 1) x(x + 1)(x + 2)
+
+
+
...
L
...
S
...
2⋅4
2⋅4⋅6
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Comparing with
2
3
1 (x)(x − 1) 1 x(x − 1)(x − 2) 1
n
n
n
2
= 1+ x
+
+
...
1⋅ 2 ⋅ 3
2 1⋅ 2 2
2
n
n(n − 1) 2
= 1+ ⋅ x +
x +
...
H
...
= 1 + +
x
x(x + 1) x(x + 1)(x + 2)
+
+
...
13
1⋅ 2 3
1⋅ 2 ⋅ 3
3
Comparing with (1 – x)–n
= 1 + n(x) +
n(n + 1) 2
x +
...
H
...
= R
...
S
...
Suppose that n is a natural number and I, F are respectively the integral part and
fractional part of (7 + 4 3) n , then show that
(i) I is an odd integer,
(ii) (I + F)(I – F) = 1
...
Given that (7 + 4 3) n = I + F where I is an integer and 0 < F < 1
...
sakshieducation
...
sakshieducation
...
e
...
e
...
e
...
+
n
C0 ⋅ 7 n − n C1 (7)n −1 (4 3) + n C 2 (7) n −2 (4 3) 2 −
...
= 2k, where k is a positive integer …(1)
Thus I + F + f is an even integer
...
Also since 0 < F < 1 and 0 < f < 1
⇒0
We get F + f = 1
(i
...
) I – F = f
…(2)
(i) From (1) I + F + f = 2k
⇒ f = 2k – 1, an odd integer
...
sakshieducation
...
sakshieducation
...
Find the coefficient of x6 in (3 + 2x + x2)6
...
(3 + 2x + x 2 ) = [(3 + 2x) + x 2 )]6
= 6 C0 (3 + 2x)6 + 6 C1 (3 + 2x)5 (x 2 ) + 6 C2 (3 + 2x) 4 (x 2 )2 + 6 C3 (3 + 2x)3 (x 2 )3 +
...
6
5
4
3
= ∑ 6 Cr ⋅ 36−r (2x)r + 6x 2 ∑ 5 Cr ⋅ 35−r (2x)r + 15x 4 ∑ 4 Cr ⋅ 34− r (2x)r + 20x 6 ∑ 3 Cr ⋅ 33−r (2x)r +
...
If n is a positive integer, then prove that C0 +
Sol
...
+ n =
...
+ n then
2
3
n +1
1
1
1 n
S = n C0 + ⋅ n C1 + ⋅ n C 2 +
...
+
⋅ Cn
1
2
3
n +1
Hence S =
∴ (n + 1)S = (n +1) C1 + (n +1) C2 + (n +1) C3 +
...
+
=
2
3
n +1
n +1
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2n +1 − 1
n +1
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23
...
+ Cn ⋅
2
3
4
n +1
=
(1 + x)n +1 − 1
(n + 1)x
x
x2
x3
xn
Sol
...
+ Cn ⋅
2
3
4
n +1
= n C0 +
1n
1
1 n
C1x + n C2 x 2 +
...
1! 2
2!
3
= 1+
n 1 n(n − 1) 2
x +
x +
...
(n + 1)x 1!
2!
3!
=
1 (n +1)
C1x + (n +1) C2 x 2 + (n +1) C3 x 3 +
...
+ n +1Cn +1x n +1 − 1
(n + 1)x
=
1
(1 + x) n +1 − 1
(n + 1)x
C0 + C1
x
x2
x3
xn
(1 + x) n +1 − 1
+ C2 ⋅ + C3 ⋅ +
...
Prove that C0 − C1 + C2 − C3 +
...
Take (1 − x)n 1 +
x
C C2
Cn
= (C0 − C1x + C2 x 2 − C3 x 3 +
...
+ n
x x
x
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...
sakshieducation
...
H
...
of (1) is = C0 − C1 + C2 − C3 +
...
H
...
of (1)
...
H
...
of (1) = (1 − x)n 1 +
x
n
(1 − x 2 ) n
1+ x
= (1 − x)
=
xn
x
n
n
= ∑ n Cr (− x 2 ) r
...
Then from (2),
n
n
1
(1 − x)n 1 + =
x
∑ n Cr (− x 2 ) r
r =0
xn
2k
∑ 2k Cr (−x 2 )r
=
2k
= ∑ 2k C r (−1)r x 2r −2k
...
of x in
n
1
(1 − x) 1 + is
x
n
2k
Ck (−1)k = n C(n / 2) (−1)n / 2
When n is odd:
Observe that the expansion in the numerator of (2) contains only even powers of x
...
e
...
x
∴ From (1), we get
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(−1)n / 2 n Cn / 2 , if n is even
2
2
2
2
2
C0 − C1 + C2 − C3 +
...
Find the coefficient of x12 in
Sol
...
(1 − 4x) 4
∞
1 + 3x
= (1 + 3x)(1 − 4x)−4 = (1 + 3x) ∑
(1 − 4x)4
r =0
(n + r −1)
Cr ⋅ X r
Here X = 4x, n = 4
∞
= (1 + 3x) ∑
r =0
∞
= (1 + 3x) ∑
r =0
(4+ r −1)
(r +3)
Cr ⋅ (4x)r
Cr ⋅ (4) r (x) r
∴ The coefficient of x12 in
1 + 3x
is
(1 − 4x) 4
= (1) ⋅ (12+3) C12 ⋅ 412 + 3 ⋅ (11+3) C3 ⋅ 411
= 15C3 ⋅ 412 + 3 ⋅ 14 C3 ⋅ 411
= 455 × 412 + (1092)411 = 728 × 412
26
...
Sol
...
+ [p + (r − 1)q] x
Tr +1 =
(r)!
q
Here X = 3x, p = 2, q = 5, r = 6,
T 6 +1 =
T7 =
r
X 3x
=
q
5
( 2 ) ( 2 + 5 ) ( 2 + 2
...
[ 2 + ( 6 − 1) 5 ] 3 x
6!
5
( 2 ) ( 7 ) (1 2 )
...
sakshieducation
...
sakshieducation
...
(27) 3
is =
6!
5
6
2
3
2 1 2 ⋅5 1 2⋅5⋅8 1
27
...
∞
3 2 3⋅ 6 2 3⋅ 6 ⋅ 9 2
2
3
2 ⋅5 1 2 ⋅5⋅8 1
+
+
...
Let S = 1 + ⋅ +
2
3
2 1 2 ⋅5 1 2 ⋅5⋅8 1
= 1+ ⋅ +
+
+
...
= (1 − x)
1! q
1⋅ 2 q
Here p = 2, q = 3,
x 1
3 1
= ⇒x= =
q 6
6 2
1
= (1 − x)− p / q = 1 −
2
−2 / 3
= 22 / 3 = 3 4
28
...
Let S =
3⋅5
3⋅5⋅ 7
3⋅5⋅ 7 ⋅9
+
+
+
...
5 ⋅10 5 ⋅10 ⋅15 5 ⋅10 ⋅15 ⋅ 20
3 ⋅5 1
1⋅2 5
2
3
+
Add 1 + 3 ⋅
3 ⋅5 ⋅7 1
3 ⋅5 ⋅7 ⋅9 1
+
1⋅2 ⋅3 5
1⋅2 ⋅3 ⋅4 5
+
...
2
+
...
sakshieducation
...
sakshieducation
...
If x =
1 1⋅ 3
1⋅ 3 ⋅ 5
2
+
+
+
...
5 5 ⋅10 5 ⋅10 ⋅15
Sol
...
5 5 ⋅10 5 ⋅10 ⋅15
2
3
1 1⋅ 3 1 1⋅ 3 ⋅ 5 1
= +
+
+
...
5 1⋅ 2 5 1⋅ 2 ⋅ 3 5
2
= 1+
3
p 1 p(p + q) 1 p(p + q)(p + 2q) 1
− p/q
+
+
= (1 − x)
1! 5
2! 5
3!
5
Here p = 1, q = 2,
2
= 1 −
5
⇒ 1+ x =
−1/2
x 1
2
= ⇒x=
q 5
5
3
=
5
−1/2
=
5
3
5
⇒ 3(1 + x)2 = 5
3
⇒ 3x 2 + 6x + 3 = 5 ⇒ 3x 2 + 6x = 2
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30
...
Sol
...
5)6
1
1 1
6
6 (0
...
5)12 +
...
0026041] = 2[0
...
9947918 = 1
...
If |x| is so small that x2 and higher powers of x may be neglected, then find an approximate
−4
3x
1/ 3
1 + (8 + 9x)
2
values of
...
(1 + 2x)2
−4
1/ 3
3x 9
= 1 + 8 1 + x
2 8
−4
(1 + 2x)−2
1/ 3
3x
9
= 1 + ⋅ 81/ 3 1 + x
2
8
(1 + 2x)−2
4 3x 1 9x
= 2 1 − 1 + [1 + (−2)(2x)]
1 2 3 8
∵ x 2 and higher powers of x are neglecting
3x
= 2(1 − 6x) 1 + (1 − 4x)
8
3x
= 2 1 − 6x + (1 − 4x)
8
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(∵ x 2 and higher powers of x are neglecting)
45
45
= 2 1 − x (1 − 4x ) = 2 1 − 4x − x
8
8
∵ x 2 and higher powers of x are neglecting
77
= 2 1 − x
8
−4
3x
1/ 3
1 + (8 + 9x)
2
77
∴
= 2 1 − x
2
8
(1 + 2x)
32
...
4
Sol
...
sakshieducation
...
sakshieducation
...
Suppose that x and y are positive and x is very small when compared to y
...
y+x
3/ 4
y
−
y+x
y
=
x
y 1 +
y
x
= 1 +
y
−3 / 4
3/ 4
3/ 4
y
−
y+x
4/5
...
1⋅ 2
y
4 y
−4 −4
−4 x 5 5 − 1 x 2
+
...
sakshieducation
...
sakshieducation
...
Expand 5 5 in increasing power of
Sol
...
5
−3 / 2
−3/2
3 5 3
3
3 5
⋅
...
∞
= 1+ +
+
...
(2r − 1) 4
+
+
...
2
1!2 5 2!2 5
r!2r
5
35
...
∞
6 ⋅12 6 ⋅12 ⋅18 6 ⋅12 ⋅18 ⋅ 24
Sol
...
6 ⋅12 6 ⋅12 ⋅18 6 ⋅12 ⋅18 ⋅ 24
2
3
4
2 ⋅ 5 1 2 ⋅ 5 ⋅ 8 1 2 ⋅ 5 ⋅ 8 ⋅11 1
⇒ 2S =
+
+
+
...
16
1 6 1⋅ 2 6 1⋅ 2 ⋅ 3 6
2
3
4
2 1 2 ⋅5 1 2 ⋅5 ⋅8 1
⇒ + 2S = 1 + +
+
+
...
1q
1⋅ 2 q
Here p = 2, q = 3,
x 1
q 3 1
= ⇒x= = =
q 6
6 6 2
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4
1
∴ + 2S = (1 − x)− p / q = 1 −
3
2
1
=
2
−2 / 3
−2 / 3
= (2)2 / 3 = 3 4
3
4
4 2
1 2
∴ 2S = 4 − ⇒ S =
− =3 −
3
2 3
2 3
3
∴
5
5 ⋅8
5 ⋅ 8 ⋅11
1 2
+
+
+
...
If the coefficients of x 9 , x10 , x11 in the expansion of (1 + x ) are in A
...
then prove that
n
n 2 − 41n + 398 = 0
...
Sol:
Given coefficients of x 9 , x10 , x11 in the expansion of (1 – x)n are in A
...
, then
2( n C10 ) = n C9 + n C11
⇒2
n!
n!
n!
=
+
(n − 10)!10! (n − 9)!9! (n − 11)!+ 11!
⇒
2
1
1
=
+
10(n − 10) (n − 9)(n − 10) 11× 10
⇒
2
110 + (n − 9)(n − 10)
=
(n − 10)10 110(n − 9)(n − 10)
⇒ 22(n − 9) = 110 + n 2 − 19n + 90
⇒ n 2 − 41n + 398 = 0
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37
...
Sol:
Number of terms in the expansion of (51/ 6 + 21/ 8 )100 are 101
...
∴ General term in the expansion of (51/ 6 + 21/ 8 )100 is
Tr +1 = 100 Cr ⋅ (51/ 6 )100−r ⋅ (21/ 8 )r
=
100
Cr
100− r
⋅5 6
r
8
⋅2
For Tr+1 to be a rational
...
∴ r = 16, 40, 64, 88
...
∴ Number of irrational terms are 101 – 4 = 97
...
If t = +
Sol:
4
...
6
...
∞ , then prove than 9t = 16
...
10 5
...
15
Given
t=
4 4
...
6
...
∞
5 5
...
10
...
6
4
...
8
+
+
+
...
10 5
...
15
⇒ 1+ t = 1+
4 1 4
...
6
...
(1)
1! 5 2! 5
3! 5
2
3
We know that
2
p x p(p + q) x
1+ +
+
1! p
2! p
3
p(p + q)(p + 2q) x
−p / q
+
...
sakshieducation
...
sakshieducation
...
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Title: 2a notes for 12th standard
Description: its very to read and understand
Description: its very to read and understand