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April 18, 2014

April 18, 2014

Posted by **ziya** on Tuesday, November 27, 2012 at 11:00am.

Sn=n(n-1)/(3)

Sn=n(2n-1)/(3)

Sn=n(n+1)/(3)

Sn=n(n+1)(2n+1)/(3)

- pre calc -
**Anonymous**, Tuesday, November 27, 2012 at 11:16amsince you are adding up n squares, the sum will be something on the order of n*n^2, or n^3. So, the first 3 choices can be eliminated right away.

Going with the last one, we see right off there's a typo.

S1 = 1 = 1(2)(3)/6

Now, assume that

1+4+9...+k^2 = k(k+1)(2k+1)/6

adding in (k+1)^2 to both sides, we get

1+4+9...+k^2+(k+1)^2 = k(k+1)(2k+1)/6 + (k+1)^2

= (2k^3+3k^2+k)/6 + (k^2+2k+1)

= (2k^3 + 3k^2 + k + 6k^2 + 12k + 6)/6

= (2k^3+6k^2+6k+2 + 3k^2+6k+3 + k+1)/6

= (2(k+1)^3 + 3(k+1)^2 + (k+1))/6

ta-da!

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