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Partial sum of n^2

  1. Nov 7, 2006 #1
    Is there a nice formula for calculating the partial sum of the series n^2 from 1 to k?
     
  2. jcsd
  3. Nov 7, 2006 #2

    quasar987

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    k(k+1)(2k+1)/6
     
  4. Nov 7, 2006 #3

    StatusX

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    You can derive that as follows:

    [tex]k^3=\sum_{n=1}^k (n^3-(n-1)^3)=\sum_{n=1}^k (n^3-(n^3-3n^3+3n-1))=3\sum_{n=1}^k n^2-3\sum_{n=1}^k n+\sum_{n=1}^k 1=3\sum_{n=1}^k n^2-3k(k+1)/2+k[/tex]

    [tex]\sum_{n=1}^k n^2=\frac{1}{3} (k^3+3k(k+1)/2-k)=\frac{1}{3}(k(k+1)(k-1)+3k(k+1)/2))=k(k+1)(k+1/2)/3=k(k+1)(2k+1)/6[/tex]
     
    Last edited: Nov 7, 2006
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