If `n in N`, then `n(n^(2)-1)` is divisible by

Explanation

Let p (n) =n .`(n^(2) -1)`
For n-1
` P (1) =1. (1^(2)-1) =0=0xx24`
Which is divisible by 24
Therefore P (n) is true for n=1
LetP (n) is true n =2m-1, Where ` m in N` so n is an odd number .
` P (n) =P (2m -1)`
`=(2m-1) [(2m-1)^(2)-1]`
`=24 lambda, " where " lambda in I`
`rArr " "(2m-1)(4m^(2)-4m)=24 lambda`
`rArr " "4m (m-1)(2m-1)=24 lambda`
For n= 2m+1
`P (2m+1) =(2m+1)[(2m+1)^(2)-1]`
`=(2m+1)(4m^(2)+4m)`
`=4m(m+1)(2m+1)`
`=4m[2m^(2)+3m+1]`
`4m[(2m^(2)-3m+1)+6m]`
`4m [(m-1)(2m-1)+6m]`
`4m(m-1)(2m-1)+24m^(2)`
`=4m(m-1)(2m-1)+24m^(2)`
`=24 lambda +24 m^(2)`
`=24 (lambda+m^(2))`
`rArr` P (n) is also true for n =2m +1.
Hence by the principle of mathematical induction the given statement P (n) is true for all odd positive integers 'n'