For a positive integer n, n^{3} + 2n is alway | vedclass

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(A) Let $f(n) = n^{3} + 2n$. We can rewrite this as $f(n) = n^{3} - n + 3n = n(n^{2} - 1) + 3n = (n-1)n(n+1) + 3n$. Here, $(n-1)n(n+1)$ is the product

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$3$

Vedclass · Answer

(A) Let $f(n) = n^{3} + 2n$. We can rewrite this as $f(n) = n^{3} - n + 3n = n(n^{2} - 1) + 3n = (n-1)n(n+1) + 3n$. Here, $(n-1)n(n+1)$ is the product of three consecutive integers, which is always divisible by $3! = 6$. However, the expression $n^{3} + 2n$ is specifically divisible by $3$ because $n^{3} \equiv n \pmod{3}$ by Fermat's Little Theorem, so $n^{3} + 2n \equiv n + 2n = 3n \equiv 0 \pmod{3}$. Testing values: For $n=1$, $1^{3} + 2(1) = 3$ (divisible by $3$). For $n=2$, $2^{3} + 2(2) = 8 + 4 = 12$ (divisible by $3$). For $n=3$, $3^{3} + 2(3) = 27 + 6 = 33$ (divisible by $3$). Thus, it is always divisible by $3$.

For a positive integer $n$, $n^{3} + 2n$ is always divisible by

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