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$n^5-5 n^3+4 n$ is divisible by 120 is true for
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The correct answer is:
all positive integers $n$
Since, $n^5-5 n^3+4 n=n\left(n^4-5 n^2+4\right)$
$=n\left(n^4-4 n^2-n^2+4\right)=n\left(n^2-1\right)\left(n^2-4\right)$
$=(n-2)(n-1)(n)(n+1)(n+2)=P(n)$ (let)
For $n=1,2, P(n)$ is zero so it is divisible by 120 . and for $n=3, P(3)=5 !=120$ and so on, so the $P(n)$ is divisible by 120 for all positive integers $n$.
$=n\left(n^4-4 n^2-n^2+4\right)=n\left(n^2-1\right)\left(n^2-4\right)$
$=(n-2)(n-1)(n)(n+1)(n+2)=P(n)$ (let)
For $n=1,2, P(n)$ is zero so it is divisible by 120 . and for $n=3, P(3)=5 !=120$ and so on, so the $P(n)$ is divisible by 120 for all positive integers $n$.
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