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1/1*2*3+1/2*3*4+……1/n(n+1)(n+2)=?

1、可以用公式求和 n(n+1)=n²+n 1*2+2*3+3*4+……+n(n+1) =1+2²+3²+…+n²+1+2+3+…+n =n(n+1)(2n+1)/6+n(n+1)/2 =n(n+1)(n+2)/3 2、可以用裂项求和 n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3 1*2+2*3+3*4+……+n(n+1) =[(1*2*3-0*1...

2/1*2+2/2*3+2/3*4+…+2/n*(n+1)= 2(1-1/2+1/2-1/3+1/3-1/4+……+1/n-1/(n+1))=2(1-1/(n+1))=2n/(n+1)

=1^2+1+2^2+2+3^2+3+....+n^2+n =(1^2+2^2+....+n^2)+(1+2+3+...+n) =n(n+1)(2n+1)/6+n(n+1)/2 =n(n+1)[(2n+1)/6+1/2] =n(n+1)(n+2)/3

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

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

1*2+2*3+3*4+...n*(n+1) =1(1+1)+2(2+1)+3(3+1)+···+n(n+1) =1²+1+2²+2+3²+3+····+n²+n =(1+2+3+····+n)+(1²+2²+3²+···n²) =(1+n)n/2+n(n+1)(2n+1)/6 =n(n+1)/2[1+(2n+1)/3] =n(n+1)(n+2)/3 此题应用的...

因为(n+1)^3-n^3=3n^2+3n+1 将n=1,2,3,.....分别代入上式可得 2^3-1^3=3x1^2+3x1+1 3^3-2^3=3x2^2+3x2+1 ...... (n+1)^3-n^3=3n^2+3n+1 将上式累加起来可得 (n+1)^3-1^3=3(1^2+2^2+3^2+....+n^2)+3(1+2+3+.....+n)+n 又1^2+2^2+3^2+....+n^2=...

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