If $n$ is odd show that $n (n^{2} - 1)$ is divisible by $24$ .
We have $n (n^{2} - 1) = n (n - 1) (n + 1)$ .

Open in App

Solution

Since $n$ is odd, $n - 1$ and $n + 1$ are two consecutive even numbers; hence one of them is divisible by $2$ and the other by $4$ .
Again $n - 1, n, n + 1$ are three consecutive numbers; hence one of them is divisible by $3$ . Thus the given expression is divisible by the product of $2, 3$ and $4$ , that is, by $24$ .

Suggest Corrections

Similar questions

n (n^{2} - 1) (3 n + 2)

24

n \in N

n (n^{2} - 1)

(n - \frac{1}{n}) (n + \frac{1}{n}) (n^{2} + \frac{1}{n^{2}})

(n - \frac{1}{n}) (n + \frac{1}{n}) (n^{2} + \frac{1}{n^{2}})

If the $n^{t h}$ term of the sequence is given by $\frac{(n^{3} + 1)}{(n + 1)}$ , then the sum of first n terms of the sequence is given by

Join BYJU'S Learning Program