s=1x2+2x3+3x4+.nx(n+1)的和为多少
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s=1x2+2x3+3x4+.nx(n+1)的和为多少
![s=1x2+2x3+3x4+.nx(n+1)的和为多少](/uploads/image/z/19556700-60-0.jpg?t=s%3D1x2%2B2x3%2B3x4%2B.nx%28n%2B1%EF%BC%89%E7%9A%84%E5%92%8C%E4%B8%BA%E5%A4%9A%E5%B0%91)
s=1x2+2x3+3x4+.nx(n+1)
=1x(1+1)+2x(2+1)+3x(3+1)+...+nx(n+1)(去括号)
=1²+1+2²+2+3²+3+...+n²+n
=(1²+2²+3²+...+n²)+(1+2+3+...+n)
下面的步骤可以套公式了;
=n(n+1)(2n+1)/6 +(1+n)*n/2
=n(n+1)(n+2)/3
=1x(1+1)+2x(2+1)+3x(3+1)+...+nx(n+1)(去括号)
=1²+1+2²+2+3²+3+...+n²+n
=(1²+2²+3²+...+n²)+(1+2+3+...+n)
下面的步骤可以套公式了;
=n(n+1)(2n+1)/6 +(1+n)*n/2
=n(n+1)(n+2)/3
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