如果1²+2²+3²+...n²=n(n+1)(2n+1)/6,那么1²+2²...+10²=?
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如果1²+2²+3²+...n²=n(n+1)(2n+1)/6,那么1²+2²...+10²=?
如果1²+2²+3²+...n²=n(n+1)(2n+1)/6,那么1²+2²...+10²=?
如果1²+2²+3²+...n²=n(n+1)(2n+1)/6,那么1²+2²...+10²=?
1²+2²...+10²
=10×(10+1)×(2×10+1)/6
=10×11×21/6
=385
1²+2²...+10²=10×11×21÷2=1155
把n=10带入
1^2+2^2+...+10^2
=10*(10+1)*(20+1)/6
=10*11*21/6
=
385