Prove the relation: 1² + 2² + 3² ... n² = 1/6 n(n+1)(2n+1)

Expert Answers

An illustration of the letter 'A' in a speech bubbles

The relation to be proved is : 1^2 + 2^2 + 3^2 ... n^2 = 1/6 n(n+1)(2n+1).

This can be done using mathematical induction.

For n = 1

S1 = 1^2 = 1 and  1*(1 + 1)(2*1 + 1)/6 = 2*3/6 = 1

Let the relation be true for n.

So we have Sn = 1^2 + 2^2 + 3^2 ... n^2 = 1/6 n(n+1)(2n+1)

Now S(n + 1) = (1/6)n(n+1)(2n+1) + (n + 1)^2

=> (n + 1)[n*(2n + 1)/6 + n + 1]

=> (1/6)(n + 1)[ 2n^2 + n + 6n + 6]

=> (1/6)(n + 1)[2n^2 + 7n + 6]

=> (1/6)(n + 1)[2n^2 + 4n + 3n + 6]

=> (1/6)(n + 1)[2n(n + 2) + 3(n + 2)]

=> (1/6)(n + 1)(n + 1 + 1)(2(n + 1) + 1)

We now have the relation true for n = 1 and if the relation is true for any n greater than 1, it it also true for n + 1.

This proves that for all values of n 1^2 + 2^2 + 3^2 ... n^2 = 1/6 n(n+1)(2n+1).

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial