The set {n(n+1)(2n+1) : n in mathbb{Z}} is a | vedclass

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(A) Let $f(n) = n(n+1)(2n+1)$.We know that $n(n+1)(2n+1) = n(n+1)(n-1+n+2) = n(n+1)(n-1) + n(n+1)(n+2)$.Each term $n(n+1)(n-1)$ and $n(n+1)(n+2)$ is a

Vedclass · Accepted Answer

$\{6k : k \in \mathbb{Z}\}$

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(A) Let $f(n) = n(n+1)(2n+1)$.We know that $n(n+1)(2n+1) = n(n+1)(n-1+n+2) = n(n+1)(n-1) + n(n+1)(n+2)$.Each term $n(n+1)(n-1)$ and $n(n+1)(n+2)$ is a product of three consecutive integers.The product of three consecutive integers is always divisible by $3! = 6$.Therefore, $n(n+1)(n-1) = 6k_1$ and $n(n+1)(n+2) = 6k_2$ for some integers $k_1, k_2$.Thus, $f(n) = 6(k_1 + k_2) = 6k$ for some integer $k$.Hence, the set is a subset of $\{6k : k \in \mathbb{Z}\}$.

The set $\{n(n+1)(2n+1) : n \in \mathbb{Z}\}$ is a subset of:

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