If n is a positive even integer,then n(n+1)(n | vedclass

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(D) The expression is $P(n) = n(n+1)(n+2)$.Since $n$ is a positive even integer,the smallest value is $n=2$.For $n=2$,$P(2) = 2 	imes 3 	imes 4 = 24

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$24$

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(D) The expression is $P(n) = n(n+1)(n+2)$.Since $n$ is a positive even integer,the smallest value is $n=2$.For $n=2$,$P(2) = 2 	imes 3 	imes 4 = 24$.For $n=4$,$P(4) = 4 	imes 5 	imes 6 = 120$.For $n=6$,$P(6) = 6 	imes 7 	imes 8 = 336$.We observe that $24$,$120$,and $336$ are all divisible by $24$.In general,the product of three consecutive integers $n(n+1)(n+2)$ is always divisible by $3! = 6$. Since $n$ is even,one of the factors must be a multiple of $4$ and another must be even,ensuring divisibility by $8$. Thus,the product is divisible by $6 	imes 4 = 24$.

If $n$ is a positive even integer,then $n(n+1)(n+2)$ is divisible by $\ldots \ldots \ldots \ldots .$

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