$If P (n) :^{''} n^{3} - n$ is divisible by $k$ for all $n > 1^{''}$ is true, then the value of $k$ is

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Solution

The correct option is B $6$
$n^{3} - n = n (n^{2} - 1) = (n - 1) n (n + 1)$
$So, n^{3} - n$ is equal to the product of three consecutive natural numbers.
$∴ n$ is divisible by $3! = 6$

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If P(n): ′′n3−n is divisible by k for all n>1′′ is true, then the value of k is

The correct option is B 6 n3−n=n(n2−1)=(n−1)n(n+1) So, n3−n is equal to the product of three consecutive natural numbers. ∴n is divisible by 3!=6

$If P (n) :^{''} n^{3} - n$ is divisible by $k$ for all $n > 1^{''}$ is true, then the value of $k$ is

The correct option is B $6$
$n^{3} - n = n (n^{2} - 1) = (n - 1) n (n + 1)$
$So, n^{3} - n$ is equal to the product of three consecutive natural numbers.
$∴ n$ is divisible by $3! = 6$