Let `(P):n(n^(n)+5)` is divisible by 6, for each natural number n.
Step I We observe that P(1) is true.
`P(1):1(1^(2)+5)=6`, which is divisible by 6.
Step II Now, assume that P(n) is true for n=k.
`P(k):k(k^(2)+5)` is divisible by 6.
`:.k(k^(2)+5)=6q`
Step III Now, to prove p(k+1) is true, we have
`P(k+1) : (k+1) [(k+1)^(2)+5]`
`=(k+1)[k^(2)+2k+1+5]`
`=(k+1)[k^(2)+2k+1+5]`
`=(k+1)[k^(2)+2k+6]`
`=k^(3)+2k^(2)+6k+k^(2)2k+6`
`=k^(3)+3k^(2)+8k+6`
`=k^(3)+5k+3k^(2)+3k+6`
`=k(k^(2)+5)+3(k^(2)+k+2)`
`=(6q)+3(k^(2)+k=2)`
We know that, `k^(2)+k+2` is divisible by 2, where, k is even or odd.
Since, `P(k+1):6q+3(k^(2)+k+2)` is divisible by 6. So, P(k+1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction P(n) is true.
