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1*2+2*3+3*4+...+n(n+1)=n(n+1)(n+2)/3

这是一个很有规律的数列求和 1*2*3=(1*2*3*4-0*1*2*3)/(4-0),括号里1*2*3是公因数,提出后剩下(4-0),把它除掉就是1*2*3了 2*3*4=(2*3*4*5-1*2*3*4)/(5-1), 同理 ... n*(n+1)(n+2)=[n*(n+1)(n+2)(n+3)-(n-1)n*(n+1)(n+2)]/[(n+3)-(n-1)] 分母都...

解法一: 1×2+2×3+3×4+...+n(n+1) =⅓×[1×2×3-0×1×2+2×3×4-1×2×3+3×4×5-2×3×4+...+n(n+1)(n+2)-(n-1)n(n+1)] =⅓n(n+1)(n+2) 解法二: 考察一般项第k项,k(k+1)=k²+k 1×2+2×3+3×4+...+n(n+1) =(1²+2²+3²+...+n...

解释过程: S=1+2+3+...+n ① S=n+(n-1)+...+1② ①+② 2S = (n+1)+(n+1)+...+(n+1) =n(n+1) S=n(n+1)/2 1+2+3+...+n=S=n(n+1)/2 这是一个等差数列的求和公式。 扩展资料等差数列是指从第二项起,每一项与它的前一项的差等于同一个常数的一种数...

1/1*2+1/2*3+1/3*4+...+1/n(n+1) =(1/1)-(1/2)+(1/2)-(1/3)+(1/3)-(1/4)+...+(1/n)-[1/(n+1)] =1-1/(n+1) =(n+1-1)/(n+1) =n/(n+1).

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...

n(n+1)(n+2)=n^3+3n^2+2n 1^3+2^3+......+n^3 =[n(n+1)/2]^2 1^2+2^2+.....+n^2=n(n+1)(2n+1)/6 1+2+3+...+n=n(n+1)/2 1*2*3+2*3*4+n(n+1)(n+2)=[n(n+1)/2]^2 +n(n+1)(2n+1)/2+n(n+1) =n(n+1)/2(n(n+1)/2+2n+3)=n(n+1)(n^2+5n+6)/4=n(n+1)(n+2)(...

n*(n+1)*(n+2)=1/4*n*(n+1)*(n+2)[n+3-(n-1)] Sn=1*2*3+2*3*4+3*4*5+...+n*(n+1)*(n+2) =1/4{1*2*3*(4-0)+2*3*4*(5-1)+3*4*5*(6-2)...+n*(n+1)*(n+2)[n+3-(n-1)]} 如此裂项相消 原式= n*(n+1)*(n+2)*(n+3)/4

解法一:(裂项相消) 因 n(n+1) = 1/3[n(n+1)(n+2) -(n-1)n(n+1)] 所以 1*2+2*3+3*4......+N*(N+1) = 1/3(1*2*3 - 0) + 1/3(2*3*4 - 1*2*3)+1/3(3*4*5 - 2*3*4) +....+ 1/3[n(n+1)(n+2) -(n-1)n(n+1)] = 1/3[(1*2*3 - 0)+(2*3*4 - 1*2*3)+(3*4*...

1*2+2*3+3*4+…+n*(n+1) =(1+2+...+n)+(1+4+9+..+n*n) =n(n+1)/2+n(n+1)(2n+1)/6 =n(n+1)(n+2)/3

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