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求和: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...

n*(n+1)=n*n+n 1*1+2*2+...n*n= n(n+1)(2n+1)/6 原式=(1+2+....+n)+(1*1+2*2+...n*n) =n(n+1)(2n+1)/6 + n(n+1)/2

解: 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)...

1x2+2x3+3x4+...+nx(n+1) = n(n+1)(n+2) /3 1x2x3+2x3x4+...+n(n+1)(n+2) = n(n+1)(n+2)(n+3) /4

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

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

令Sn=x^2+2x^3+3x^4+...+nx^(n+1) xSn=x^3+2x^4+3x^5+...+nx^(n+2) 上式减下式,得: (1-x)Sn=x^2+x^3+x^4+...+x^(n+1)-nx^(n+2) =x^2*(1-x^n)/(1-x)-nx^(n+2) =[x^2-x^(n+2)-nx^(n+2)+nx^(n+3)]/(1-x) =[x^2-(n+1)x^(n+2)+nx^(n+3)]/(1-x) Sn=[...

解: N(N+1)(N+2)= N(N+1)(N+2)[N+3-(N-1)]/4=[N(N+1)(N+2)(N+3)-(N-1)N(N+1)(N+2)]/4 则原式= 1/4×[1*2*3*4-0*1*2*3+2*3*4*5-1*2*3*4+3*4*5*6-2*3*4*5+-------N(N+1)(N+2)(N+3)-(N-1)N(N+1(N+2)] =N(N+1)(N+2)(N+3)/4

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...

Sn=1+2x+3x^2+4x^3+5x^4……+nx^n-1 两端同乘 x x * Sn = x + 2x^2 + 3x^3 + 4x^4 + …… +(n-1)*x^(n-1) + nx^n 两式相减 Sn - x Sn = 1 + (2x -x) + (3x^2 - 2x^2) + (4x^3 - 3x^3) + …… +[nx^(n-1) - (n-1)*x^(n-1)] - nx^n (1-x)Sn = 1 + x + x^...

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