If n is a positive integer,then n^{3}+2n is d | vedclass

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(D) Let $P(n) = n^{3} + 2n$. For $n = 1$,$P(1) = 1^{3} + 2(1) = 1 + 2 = 3$. For $n = 2$,$P(2) = 2^{3} + 2(2) = 8 + 4 = 12$. For $n = 3$,$P(3) = 3^{3}

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Let $P(n) = n^{3} + 2n$. For $n = 1$,$P(1) = 1^{3} + 2(1) = 1 + 2 = 3$. For $n = 2$,$P(2) = 2^{3} + 2(2) = 8 + 4 = 12$. For $n = 3$,$P(3) = 3^{3} + 2(3) = 27 + 6 = 33$. We observe that $3, 12, 33$ are all divisible by $3$. Alternatively,$n^{3} + 2n = n^{3} - n + 3n = n(n^{2} - 1) + 3n = (n-1)n(n+1) + 3n$. Since $(n-1)n(n+1)$ is the product of three consecutive integers,it is always divisible by $3! = 6$. However,the expression $n^{3} + 2n$ is always divisible by $3$ for any positive integer $n$.

If $n$ is a positive integer,then $n^{3}+2n$ is divisible by

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