1²+2²+3²+...n²=1/6n(n+1)(2n+1)),试求(1)11²+12²+14²+...+20²
人气:143 ℃ 时间:2020-02-15 08:53:17
解答
11²+12²+14²+...+20²
=1^2+2^2+.+20^2-(1^2+2^2+...+10^2)
=1/6*20(20+1)(2*20+1)-1/6*10*(10+1)(2*10+1)
=1/6*(20*21*41-10*11*21)
=210/6(410-11)
=35*399
=13965不好意思,第二步不懂,能详细讲解一下吗,谢谢
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