prove that `n(n^(2)+5)` is divisible by 6, for each natural number n.

1.	prove that `n(n^(2)+5)` is divisible by 6, for each natural number n.
Answer» 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.

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