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The sum 1^2 + 2^2 + 3^2 + ... + n^2 = n(n + 1)(2n + 1)/6.

 

What is the value of 21^2 + 22^2 + 23^2 + ... + 40^2 + 41^2 + 42^2?

 Apr 4, 2021
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To solve this, we need to find 1^2...42^2 and then subtract 1^2...20^2 from it. 

You already know the formula 1^2 + 2^2 + 3^2 + ... + n^2 = n(n + 1)(2n + 1)/6.

 

42(42+1)(2*42+1)/6-20(20+1)(2*20+1)/6 = 22715

 

I hope this helped. :))

 

=^._.^=

 Apr 4, 2021

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