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Friction

November 15, 2015

Praveen Fernando

1

Mathematical Induction

Apart from normal methods that we are using to solve mathematical equations or expressions there exists
some other methods
...

In this chapter we are comprehensively studying mathematical induction and it’s uses
...

Let(xy)n = xn yn be a mathematical expression which can be represented by Pn where n ∈ R+
...

Hence,
when n=1; (xy)1 = xy is represented by P1
when n=2; (xy)2 = x2 y 2 is represented by P2
when n=3; (xy)3 = x3 y 3 is represented by P3

...
But for all infinite number of n we can’t prove that Pn from the above method
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2

Mathematical Induction Concept

Let us consider an expression consists of the n variable which can be represented by Pn
...

i)P1 is true in the above statement
...
This can be written as Pk =⇒ Pk+1
1

iii)Prove that Pk+1 is true with the use of the expression Pk
When the above i) and ii) are satisfied , Pn is said to be true for all n positive integers
...

Therefore to show that an expression, Pn is true for all n(∈ R+ ) the following steps will be followed
...

ii) :Assuming that Pk is true for all k
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3

Solved Examples

3
...

2

Proof:
The summation of first n numbers can be written as follows:
1 + 2 + 3 + 4 + 5 +
...
+ n =

n(N + 1)
2

for n = 1; P1 : 1 = 1(1 + 1)/2 = 1
∴ P1 is true
Now let us assume that the given expression is true for n = k
...
+ k =

k(k + 1)
− − − − − − − − − −(1)
2

Now let us prove that the expression is true for n = k + 1
...
+ k + (k + 1) =

k(k + 1)
+ (k + 1)
2

= (k + 1)((k/2) + 1)
=
1 + 2 + 3 + 4 +
...

∴ Pn is true f or all n(∈ R+ )

2

3
...
+ (2n − 1) = n2
Proof:
W hen n = 1;
L
...
S = 1
R
...
S = 1 ∗ 1 = 1
∴ L
...
S = R
...
S
...
T hen :
1 + 3 + 5 + 7 +
...
+ (2k − 1) + [2(k + 1) − 1] = k 2 + [2(k + 1) − 1]
1 + 3 + 5 +
...
+ [2(k + 1) − 1] = (k + 1)2
∴ Pn is true f or n = k + 1
So P1 is true
and Pk + 1 is true when Pk is assumed to be true
...
3

Example

W hen n is a positive number show that,
5
6
7
n+4
n(3n + 7)
+
+

...
2
...
3
...
4
...
H
...
2
...
H
...
2
...
+
=
− − − − − − − − − − − (1)
1
...
3 2
...
4 3
...
5
k(k + 1)(k + 2)
2(k + 1)(k + 2)
Ading

(k + 1) + 4
to above (1) yields;
(k + 1)[(k + 1) + 1][(k + 1) + 2]

3

5
6
7
k+4
(k + 1) + 4
k(3k + 7)
(k + 5)
+
+

...
2
...
3
...
4
...
+
+
=
[
+
]
1
...
3 2
...
4 3
...
5
k(k + 1)(k + 2) (k + 1)[(k + 1) + 1][(k + 1) + 2]
(k + 1)(k + 2)
2
k+3
6
7
k+4
(k + 1) + 4
3k 3 + 16k 2 + 23k + 10
5
+
+

...
2
...
3
...
4
...
+
+
=
1
...
3 2
...
4 3
...
5
k(k + 1)(k + 2) (k + 1)[(k + 1) + 1][(k + 1) + 2]
2(k + 1)(k + 2)(k + 3)
5
6
7
k+4
(k + 1) + 4
(k + 1)(3(k + 1) + 7)
+
+

...
2
...
3
...
4
...

∴ Pn is true f or all n(∈ R+ )

3
...
+ (2n − 1) = n2
Proof:
W hen n = 1;
L
...
S = 1
R
...
S = 1 ∗ 1 = 1
∴ L
...
S = R
...
S
...
T hen :
1 + 3 + 5 + 7 +
...
+ (2k − 1) + [2(k + 1) − 1] = k 2 + [2(k + 1) − 1]
1 + 3 + 5 +
...
+ [2(k + 1) − 1] = (k + 1)2
∴ Pn is true f or n = k + 1
So P1 is true
and Pk + 1 is true when Pk is assumed to be true
...
5

Example

Show that x2n − y 2n is divided by (x + y)
proof:
W hen n = 1,
x2n − y 2n
Pn =
x−y
x2∗1 − y 2∗1
P1 =
x−y
P1 = x + y(∈ R)
∴ Pn true f or n = 1
N ow let assume Pn is true f or n = k;
i
...


x2k − y 2k
= L(∈ R)
x−y

N ow let us show that Pn is true f or n = (k + 1) then,
2(k+1)

x2(k+1)−y
(x − y)

=

=

x2 (x2k ) − y 2 (y 2k )
(x − y)
x2 [x2k − y 2k ] y 2 (y 2k ) − x2 (y 2k )
−
(x − y)
(x − y)

= x2 L + y 2k

x2 − y 2
(x − y)

= [x2 L + y 2k (x + y)] (∈ R)
∴ Pn is true f or n = k + 1
So P1 is true
and Pk + 1 is true when Pk is assumed to be true

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