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

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• Aug 28th 2011, 03:27 AM
Stuu
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.
• Aug 28th 2011, 03:29 AM
anonimnystefy
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?
• Aug 28th 2011, 03:42 AM
Stuu
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
• Aug 28th 2011, 03:52 AM
anonimnystefy
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:
$11^2 + 12^2 + 13^2 + ... + n^2=1^2 + 2^2 + 3^2 + ... + n^2-(1^2+2^2+3^2+...+11^2)=\frac{n\cdot (n+1)\cdot(2n+1)}{6}-506$
• Aug 28th 2011, 03:57 AM
Plato
Re: Find the expression for 11^2 + 12^2+13^2+...+ n^2
Quote:

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

$\sum\limits_{k = 11}^n {k^2 } = \sum\limits_{k = 1}^n {k^2 } - \sum\limits_{k = 1}^{10} {k^2 }$