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04 Jul 2007, 17:25
Kudos
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Find the sum of the first 20 terms of the series:
1^2 + 2^2 + 3^2 + ... + 20^2
Does anyone know of a formula? Thanks.
1^2 + 2^2 + 3^2 + ... + 20^2
Does anyone know of a formula? Thanks.
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04 Jul 2007, 17:59
Kudos
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I'm pretty sure it's:
(n*(n+1)*(2n+1))/6
So if we had use 4, it would be (4*5*9)/6 = 180/6 = 30
1 + 4 + 9 + 16 = 30
20 would be:
(20*21*41)/6 = 2870
You may want to check that...
(n*(n+1)*(2n+1))/6
So if we had use 4, it would be (4*5*9)/6 = 180/6 = 30
1 + 4 + 9 + 16 = 30
20 would be:
(20*21*41)/6 = 2870
You may want to check that...
04 Jul 2007, 18:29
Kudos
Bookmarks
mavery
wonderful - thanks!
seems to work for some no.s i tried. can you plz point me to some definitive reference on how this formula is arrived at - couldn't find it online.
also, is there a variation of this formula that is based on first term, difference, etc. (like arithmetic series Sum formula).
For example, is there a way to solve?
1. 1^2 + 3^2 + 5^2 + ...19^2
2. 4^2 + 7^2 + 10^2 + ...25^2
