The expression n^5-5n^3+4n is divisible by 12 | vedclass

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(A) Let $P(n) = n^5-5n^3+4n$.Factoring the expression: $P(n) = n(n^4-5n^2+4) = n(n^2-1)(n^2-4)$.Further factoring: $P(n) = (n-2)(n-1)n(n+1)(n+2)$.This

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all positive integers $n$

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(A) Let $P(n) = n^5-5n^3+4n$.Factoring the expression: $P(n) = n(n^4-5n^2+4) = n(n^2-1)(n^2-4)$.Further factoring: $P(n) = (n-2)(n-1)n(n+1)(n+2)$.This is the product of $5$ consecutive integers.The product of $k$ consecutive integers is always divisible by $k!$.Therefore,$P(n)$ is divisible by $5! = 120$ for all integers $n \geq 1$.For $n=1$,$P(1) = (-1)(0)(1)(2)(3) = 0$,which is divisible by $120$.For $n=2$,$P(2) = (0)(1)(2)(3)(4) = 0$,which is divisible by $120$.For $n=3$,$P(3) = 1 	imes 2 	imes 3 	imes 4 	imes 5 = 120$,which is divisible by $120$.Thus,the statement is true for all positive integers $n$.

The expression $n^5-5n^3+4n$ is divisible by $120$ for which of the following?

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