For all n in N,if n(n^2+3) is divisible by k, | vedclass

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(D) Let $f(n) = n(n^2+3) = n^3+3n$. For $n=1$,$f(1) = 1(1+3) = 4$. For $n=2$,$f(2) = 2(4+3) = 2(7) = 14$. For $n=3$,$f(3) = 3(9+3) = 3(12) = 36$. For

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$2$

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(D) Let $f(n) = n(n^2+3) = n^3+3n$. For $n=1$,$f(1) = 1(1+3) = 4$. For $n=2$,$f(2) = 2(4+3) = 2(7) = 14$. For $n=3$,$f(3) = 3(9+3) = 3(12) = 36$. For $n=4$,$f(4) = 4(16+3) = 4(19) = 76$. We check the greatest common divisor of these values: $gcd(4, 14, 36, 76) = 2$. Thus,the expression $n(n^2+3)$ is always divisible by $2$ for all $n \in N$. Since $f(1)=4$ and $f(2)=14$,the only common divisor for all $n$ is $2$.

For all $n \in N$,if $n(n^2+3)$ is divisible by $k$,then the maximum value of $k$ is

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