If n be any natural number then by whichlargest number `(n^3-n)`is always divisible?3(b) 6 (c) 12 (d) 18

If n be any natural number then by whichlargest number `(n^3-n)`is alw

1.	If n be any natural number then by whichlargest number `(n^3-n)`is always divisible?3(b) 6 (c) 12 (d) 18
Answer» Let P(n) : `n^(3)-n` is divisible by 6, for each natural bumber `nle2`. Stwp I We observe that P(2) is true. P(2) : `(2)^(3)-2`. `rArr 8-2=6`, which is divisible by 6. Stwep II Now, assume that P(n) is true for n=k. P(k) : `k^(3)-k` is divisible by 6. `:.k^(3)-k=6q` Step III To prove P(k+1) is true `P(k+1):(k+1)^(3)-(k+1)`. `=k^(3)+1+3k(k+1)-(k+1)` `=k^(3)+1+3k^(2)+3k-k-1` `=k^(3)-k+3k^(2)+3k` `=6q+3k(k+1)` [from step II] We know that, 3k (k+1) is divisible by 6 for each natural number n=k. So, P(k+1) is true. Hence, by the principle of mathematical induction P(n) is true.