已知1²+2²+3²+…+n²=n(n+1)(2n+1)/6,试求:50²+51²+52²+…+101²

问题描述:

已知1²+2²+3²+…+n²=n(n+1)(2n+1)/6,试求:50²+51²+52²+…+101²
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50²+51²+52²+…+101²
=(1²+2²+3²+…+101²)-(1²+2²+3²+…+49²)
=101(101+1)(2*101+1)/6 -49(49+1)(2*49+1)/6
=101*102*203/6 -49*50*99/6
=308126