# Thread: Find the expression for 11^2 + 12^2+13^2+...+ n^2

1. ## Find the expression for 11^2 + 12^2+13^2+...+ n^2

Question : Write an expression for: 11^2 + 12^2 + 13^2 + ... + n^2

The previous question had the expression:
1+2^2 = n x (n+1) x (2n+1) /6
This expression is a quick way to add square numbers together starting at one. I've got to find a similar expression which adds square numbers together, but the pattern must start at 11^2.

2. ## Re: Find the expression for 11^2 + 12^2+13^2+...+ n^2

hi Stuu

11^2 + 12^2 + 13^2 + ... + n^2=1^2 + 2^2 + 3^2 + ... + n^2-(1^2+2^2+3^2+...+11^2)

can you do it from here?

3. ## Re: Find the expression for 11^2 + 12^2+13^2+...+ n^2

Is there any expression similar to; n x (n+1) x (2n+1) /6

4. ## Re: Find the expression for 11^2 + 12^2+13^2+...+ n^2

using the identity you gave you get:
1^2+2^2+3^2+...+n^2=n * (n+1) * (2n+1)/6
and
1^2+2^2+3^3+...+11^2=11*(11+1)*(2*11+1)/6=11*12*23/6=11*2*23=11*46=506

so you get:
$11^2 + 12^2 + 13^2 + ... + n^2=1^2 + 2^2 + 3^2 + ... + n^2-(1^2+2^2+3^2+...+11^2)=\frac{n\cdot (n+1)\cdot(2n+1)}{6}-506$

5. ## Re: Find the expression for 11^2 + 12^2+13^2+...+ n^2

Originally Posted by Stuu
Is there any expression similar to; n x (n+1) x (2n+1) /6
$\sum\limits_{k = 11}^n {k^2 } = \sum\limits_{k = 1}^n {k^2 } - \sum\limits_{k = 1}^{10} {k^2 }$