Prove that if n is an odd integer then 8/n^2 -1.

Theorem:

We prove it by mathematical induction

For n=1

= n^2 -1

= 1^2 -1

= 1-1

= 0

CASE-I is true for n=1

Suppose it is true for n = k where k is a odd integer.

i.e.

= 8/k^2 -1 -------------(1)

we have to prove it for n= k+2

= n^2 -1

putting n = k+2, we have

= (k+2)^2 -1

= k^2 + 4k + 4 -1

=k^2 =1 + 4k + 4

= (k^2 -1) + 4(k + 1) -------------(2)

As k is an odd integer so k+1 is even, So we know that

= 2|k+1

= 4(2)|4(k+1)

= 8|4k + 4

By eq.1 , we have

= 8|[k^2 -1 + 4(k+1)]

from eq.2, we have

= 8| n^2 -1

CASE-II is true for n=k+1

Hence It is true for every odd integer.

odd integer