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Math Help - Find the expression for 11^2 + 12^2+13^2+...+ n^2

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

    Quote Originally Posted by Stuu View Post
    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 }
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