Show that $n (n + 1) (2 n + 1)$ is multiple of $6$ for every natural number $n$ .

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Solution

Given $n (n + 1) (2 n + 1)$
If $n = 1$ then,
$1 (1 + 1) (2 (1) + 1) = 1 (2) (3) = 6$ which is a multiple of 6.
If $n = 2$ then,
$2 (2 + 1) (2 (2) + 1) = 2 (3) (5) = 30$ which is a multiple of 6.
If $n = 3$ then,
$3 (3 + 1) (2 (3) + 1) = 3 (4) (7) = 84$ which is also a multiple of 6.
Hence
$n (n + 1) (n + 1)$ is also multiple of 6 for every natural number of n.

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Show that n(n+1)(2n+1) is multiple of 6 for every natural number n.

Given n(n+1)(2n+1)If n=1 then,1(1+1)(2(1)+1)=1(2)(3)=6 which is a multiple of 6.If n=2 then,2(2+1)(2(2)+1)=2(3)(5)=30 which is a multiple of 6.If n=3 then,3(3+1)(2(3)+1)=3(4)(7)=84 which is also a multiple of 6.Hencen(n+1)(n+1) is also multiple of 6 for every natural number of n.

Show that $n (n + 1) (2 n + 1)$ is multiple of $6$ for every natural number $n$ .