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1*22+2*32+3*42+……+n(n+1)2=

因为 k*(k+1) = k² + k 所以 1*2 + 2*3 + 3*4 + ... + n*(n+1) = (1²+1) + (2²+2) + (3²+3) + ... + (n²+n) = (1²+2²+3²+...+n²) + (1+2+3+...+n) = n(n+1)(2n+1)/6 + n(n+1)/2 = [n(n+1)/6] * (2...

证明:1×2+2×3+3×4+......+n(n+1) =(1×1+1)+(2×2+2)+(3×3+3)+......(n×n+n) =(1^2+2^2+3^2+......n^2)+(1+2+3+......n) =n*(n+1)*(2*n+1)/6+n(n+1)/2 =n(n+1)(n+2)/3

n*(n+1)*(n+2) =n³+3n²+2n 1³+……+n³=n²(n+1)²/4 1²+……+n²=n(n+1)(2n+1)/6 1+……+n=n(n+1)/2 所以原式=n²(n+1)²/4+3n(n+1)(2n+1)/6+2n(n+1)/2 =n(n+1)(n+2)(n+3)/4

(n+1)^3-n^3=3n^2+3n+1 n^3-(n-1)^3=3(n-1)^2+3(n-1)+1 …… 2^3-1^3=3*1^2+3*1+1 全都加起来,左边中间可以抵消掉 (n+1)^3-1=3*[n^2+(n-1)^2+(n-2)^2+……+2^2+1^2]+3*[n+(n-1)+……+3+2+1]+n*1 而(n+1)^3-1=(n+1-1)[(n+1)^2+(n+1)+1] =n(n^2+3n+3) n+...

裂项法: 同乘以3后: 原式=1*2*3+2*3*3+3*4*3+....+(n-1)*n*3 =1*2*3+2*3*(4-1)+3*4*(5-2)+....(n-1)n*[(n+1)-(n-2)] =1*2*3+2*3*4-1*2*3+3*4*4-2*3*4+(n-1)n(n+1)-(n-2)(n-1)n =(n-1)n(n+1) 再除以3, 结果是(n-1)n(n+1)/3

#include int main(){int n,sum=0,i;printf("输入n:");scanf("%d",&n);for(i=1;i

n(n+1)=n^2+n

Private Sub Command1_Click()Dim n As IntegerDim sum As Longsum = 0n = InputBox("请输入数字N")For i = 1 To nsum = sum + i * (i + 1)Next iPrint sumEnd Sub求采纳!

求1^2+2^2+3^2+...+n^2的值(答案n(n+1)(2n+1)/6) 方法一:利用立方差公式 n^3-(n-1)^3=1*[n^2+(n-1)^2+n(n-1)] =n^2+(n-1)^2+n^2-n =2*n^2+(n-1)^2-n 2^3-1^3=2*2^2+1^2-2 3^3-2^3=2*3^2+2^2-3 4^3-3^3=2*4^2+3^2-4 ...... n^3-(n-1)^3=2*n^2+...

当n=1时,1*2=1*(1+1)*(1+2)/3, 该等式成立 现在假设n=k时,1*2+2*3+...+k(k+1)=k(k+1)(k+2)/3 成立 k为自然数 则当n=k+1时,1*2+2*3+...+k(k+1)+(k+1)(k+2) =k(k+1)(k+2)/3+(k+1)(k+2) =(k/3+1)(k+1)(k+2) =(k+1)(k+1+1)(k+1+2)/3 即,当n=k+1...

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