Determine the subset for the set {n(n+1)(2n+1 | vedclass

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(A) Let $f(n) = n(n+1)(2n+1)$. We know that $2n+1 = (n-1) + (n+2)$,so $n(n+1)(2n+1) = n(n+1)(n-1+n+2) = n(n+1)(n-1) + n(n+1)(n+2)$. Note that $n(n+1)(

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

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(A) Let $f(n) = n(n+1)(2n+1)$. We know that $2n+1 = (n-1) + (n+2)$,so $n(n+1)(2n+1) = n(n+1)(n-1+n+2) = n(n+1)(n-1) + n(n+1)(n+2)$. Note that $n(n+1)(n-1)$ is the product of three consecutive integers,which is always divisible by $3! = 6$. Similarly,$n(n+1)(n+2)$ is also the product of three consecutive integers,which is always divisible by $3! = 6$. Therefore,$f(n)$ is divisible by $6$ for all $n \in \mathbb{Z}$. Thus,$\{n(n+1)(2n+1) : n \in \mathbb{Z}\} \subset \{6k : k \in \mathbb{Z}\}$.

$(i)$ $\{1, 2, 3, 6\}$	$(a)$ $\{x : x \text{ is a prime number and a divisor of } 6\}$
$(ii)$ $\{2, 3\}$	$(b)$ $\{x : x \text{ is an odd natural number less than } 10\}$
$(iii)$ $\{M, A, T, H, E, I, C, S\}$	$(c)$ $\{x : x \text{ is a natural number and divisor of } 6\}$
$(iv)$ $\{1, 3, 5, 7, 9\}$	$(d)$ $\{x : x \text{ is a letter of the word } MATHEMATICS\}$

Determine the subset for the set $\{n(n+1)(2n+1) : n \in \mathbb{Z}\}$.

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