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1x2+2x3+3x4+...+nx(n+1)=

因为n(n+1)=n平方+n 原式=(1平方+2平方+……n平方)+(1+2+3……n) =n(n+1)(2n+1)/6+n(n+1)/2为标准答案 注:Sn(1平方+2平方+……n平方)的证明: (a+1)³-a³=3a²+3a+1(即(a+1)³=a³+3a²+3a+1) a=1时:2³...

先求它们的倒数和,再把结果倒回来。 1x2+2x3+3x4+4x5+5x6+6x7+7x8+8x9+9x10+10x11 =10×(10+1)×(10+2)÷3 =10×11×12÷3 =110×12÷3 =1320÷3 =440

1x2+2x3+3x4+…+n(n+1)  =1^2+1+2^2+2+3^2+3+…+n^2+n  =(1^2+2^2+3^2+…+n^2)+(1+2+3+…+n)  =1/6*n(n+1)(2n+1)+1/2*n(n+1)  =1/6*n(n+1)(2n+1+3)(提取公因式)  =1/3*n(n+1)(n+2)

1x2+2x3+3x4+4x5+...+n(n+1) =(1^2+2^2+……n^2)+(1+2+3+……n) =n(n+1)(2n+1)/6+(1+n)xn/2 =n(n+1)(n+2)/3

一般的,有: (n-1)n(n+1) =n^3-n {n^3}求和公式:Sn=[n(n+1)/2]^2 {n}求和公式:Sn=n(n+1)/2 1x2x3+2x3x4+3x4x5+....+7x8x9 =2^3-2+3^3-3+...+8^3-8 =(2^3+3^3+...+8^3)-(2+3+...+8) =[(8*9/2)^2-1]-8*9/2+1 =1260

1x2+2x3+3x4+…+n(n+1) =1x(1+1)+2x(2+1)+3x(3+1)+…n(n+1) =(1^2+2^2+3^2+…+n^2)+(1+2+3+…+n) =n(n+1)(2n+1)/6+n(n+1)/2 =n(n+1)[(2n+1)+3]/6 1x2+2x3+3x4+…+10x11 =10x(10+1)x[(10x2+1)+3]/6 =110x4 =440

1/2(1/1*2-1/2*3) 1/2(1/2*3-1/3*4)... 1/2[1/n*(n 1)-1/(n 1)*(n 2)] =1/2*[1/1*2-1/(n 1)(n 2)]=1/4-1/2(n 1)(n 2) 1/(n*(n 1)*(n 2))=1/2*(1/(n*(n 1))-1/((n 1)*(n 2))) 所以:原式=1/2*((1/(1*2)-1/(2*3)) (1/(2*3)-1/(3*4)) ...) =1/2*(1...

1x2+2x3+3x4+…+n(n+1) =1x(1+1)+2x(2+1)+3x(3+1)+…+n(n+1) =1²+1+2²+2+3²+3+…+n²+n =(1²+2²+3²+…+n²)+(1+2+3+…+n) =(1/6)n(n+1)(2n+1)+(1/2)*n(n+1) =(1/6)n(n+1)(2n+1+3) (提取公因式) =(1/3)n(n+1)...

因为(n-1)n(n+1)=n(n²-1)=n³-n ∴原式=2³-2+3³-3+4³-4+……+99³-99 =2³+3³+4³+……+99³-(2+3+4+……+99) =1³+2³+3³+……+99³-(1³+2+3+……+99) 【这里要知道:连续自然数的立...

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