Prove that 1^2 + 2^2+ 3^2 + · · · + n^2 =n(n + 1)(2n + 1)/6. by mathematical induction..  

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember that the algorithm of inductive method consists of two steps such that:

1. Prove the base case for n = 1, or n > m; m is a natural given number

2. Prove that the statement works for n+1 if it works for n.

You need to check if the statement works for n=1 such that:

`1^2 = (1)(1+1)(2+1)/6 =gt 1 = 6/6 =gt 1=1`

Hence, if the statement works for n=1, then you can move to the next step.

Consider that the statements works for n term and you need to prove that it works for n+1 terms.

`1^2 + 2^2 + 3^2 + ... + n^2 = (n(n+1)(2n+1))/6`

You need to prove that the statement work for n+1 such that:

`1^2 + 2^2 + 3^2 + ... + n^2 + (n+1)^2 = ((n+1)((n+1)+1)(2(n+1)+1))/6`

Substituting to the left `(n(n+1)(2n+1))/6 ` for`1^2 + 2^2 + 3^2 + ... + n^2`  yields:

`(n(n+1)(2n+1))/6 + (n+1)^2 = ((n+1)(n+2)(2n+3))/6`

`` Factoring out n+1 both sides yields:

`(n+1)(n(2n+1)/6 + n+1) = ((n+1)(n+2)(2n+3))/6`

Reducing both sides by n+1 yields:

`(n(2n+1)/6 + n+1) = ((n+2)(2n+3))/6`

Opening the brackets yields:

`(2n^2 + 7n + 6)/6 = (2n^2 + 3n + 4n + 6)/6`

`(2n^2 + 7n + 6)/6 = (2n^2 + 7n + 6)/6`

The last line proves that the statement `1^2 + 2^2 + .... + n^2 = (n(n+1)(2n+1))/6`  works for any n natural value.

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