Question

Prove that for any positive interger n, n^3 - n is divisible by 6.​

Answer 1

Let us consider

a = n3 – n

a = n(n2 – 1)

a = n(n + 1)(n – 1)

Assumtions:

1. Out of three (n – 1) , n, (n + 1) one must be even so a is divisible by 2.

2. (n – 1) , n, (n + 1) are consecutive integers thus as proved a must be divisible by 3.

From (1) and (2) a must be divisible by 2 × 3 = 6

Thus, n³ – n is divisible by 6 for any positive integer n.

Answer 2

Let p(n)=(n^3-n)/6

p(1)=(1^3–1)\6=0/6=0

p(2)=(2^3–2)/6=6/6=1

Let us assume that it is true for p(k). To prove

P(k+1) is also true

[(k+1)^3-(k+1)]\6

(k+1)[(k+1)^2–1]\6

(k+1)(k^2+2k+1–1)/6

(k+1)(k)(k+2)/6 *

• This result is similar to ( n^3-n)/6=(n-1)n(n+1)/6
• Hence it is true for all values of n