Prove that `3^(2n)+24n-1` is divisible by 32 .

Explanation

Let `P(n)=(25)^(n+1)-24n+5735`
Step I For `n=1`.
`P(1)=(25)^2-24+5735=625 -24+6336=11xx(24)^2`, which is divisible by `(24)^2`.
Therefore , the result is true for `n=1`.
Step II Assume that the result is true for `n=k`. Then , `P(k)=(25)^(k+1)-24k+5735` is divisible by `(24)^2`.
`rArr P(k)=(24)^2r`, where r is an integer .
Step III For `n=k+1`,
`P(k+1)=(25)^((k+1)+1) -24(k+1)+5735`
`=(25)^(k+2)-24k+5711`
`=(25)(25)^(k+1)-24k+5711`
Now , `P(k+1)-P(k)`
`={(25)(25)^(k+1)-24k+5711}-{(25)^(k+1)-24k+5735}`
`=(24)(25)^(k+1)-24`
`=24{(25)^(k+1)-1}`
`rArr P(k+1)=P(k)+24{(25)^(k+1)-1}`
But the assumption P(k) is divisible by `(24)^2`. Also ,`24{(25)^(k+1)-}` is clearly divisible by `(24)^2`, forall k in N`. This shows that , the result is true for `n=k+1`.
Hence , the principle of mathematical induction , result is true for all `n in N`.