If `n` be any natural number then by which largest number `(n^3-n)` is always divisible ?

Options

A. 3
B. 6
C. 12
D. 18

Answer

B

Explanation

Let P(n) : `n^(3)-n` is divisible by 6, for each natural bumber `nle2`.
Stwp I We observe that P(2) is true. P(2) : `(2)^(3)-2`.
`rArr 8-2=6`, which is divisible by 6.
Stwep II Now, assume that P(n) is true for n=k.
P(k) : `k^(3)-k` is divisible by 6.
`:.k^(3)-k=6q`
Step III To prove P(k+1) is true
`P(k+1):(k+1)^(3)-(k+1)`.
`=k^(3)+1+3k(k+1)-(k+1)`
`=k^(3)+1+3k^(2)+3k-k-1`
`=k^(3)-k+3k^(2)+3k`
`=6q+3k(k+1)` [from step II]
We know that, 3k (k+1) is divisible by 6 for each natural number n=k.
So, P(k+1) is true. Hence, by the principle of mathematical induction P(n) is true.