1²+2²+3²+…+n²= n(n+1)(2n+1)/6
So for this problem
(11² + 12² + 13² +…+21²)
[1²+2²+…+21²] - [1²+2²+3²+…+10²]
= (21*21*41)/6 - (10*11*21)/6
2871-385
=2486
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Verified answer
1²+2²+3²+…+n²= n(n+1)(2n+1)/6
So for this problem
(11² + 12² + 13² +…+21²)
[1²+2²+…+21²] - [1²+2²+3²+…+10²]
= (21*21*41)/6 - (10*11*21)/6
2871-385
=2486