Question

For any positive integer n,prove that n^3 -n is divided by 6.​

Answer 1

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

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