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

Explanation

Let `P(n)=3^(2n)+24n-1`
Step I For `n=1`,
`P(1)=3^(2xx1)+24xx1-1=3^2+24-1=9+24-1=32`, which is divisible by 32.
Therefore , the result is true for `n=1`
Step II Assume that the result is true for `n=k`, then `P(k)=3^(2k)+24k-1` is divsible by 32.
`rArr P(k)=32r`, where r is an integer .
Step III For `n=k+1`,
`P(k+1)=3^(2(k+1))+24(k+1)-1`
`=3^(2k+2)+24k+24-1`
`=3^2. 3^(2k)+24k+23`
`9.3^(2k)+24k+23`

`therefore P(k+1)=9(3^(2k)+24k-1)-32(6k-1)`
`=9P(k)-32(6k-1)`
`therefore P(k+1)=9(32r)-32(6k-1)`
`=32(9r-6k+1)`,
which is divisible by 32 , as `9r-6k+1` is an integer . Therefore , `P(k+1)` is divisible by 32. divisible by 32. Hence , by the principle of mathematical induction P(n) is divisible by 32, `forall n in N`.