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1×2×3+2×3×4+3×4×5+…+n(n+1)(n+2)=?

原式= n*(n+1)*(n+2)*(n+3)/4

这种题直接代公式就可以了 1*2*3+2*3*4+3*4*5+4*5*6++n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4= 46*C n+3表示从n+3个数中选4个 再乘以6

裂项相消n*(n+1)*(n+2) = [n*(n+1)*(n+2)*(n+3) - (n-1)*n*(n+1)*(n+2)]/4n(*n+1) = [n*(n+1)*(n+2) - (n-1)*n*(n+1)]/31*2*3+2*3*4+3*4*5+…+7*8*9 = [1*2*3*4 - 0*1*2*3 + 2*3*4*5 - 1*2*3*4 + … + 7*8*9*10 - 6*7*8*9]/4 = 7*8*9*10/4 = 12601*2+2*3+3*4+

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)]分母都是4相加,所有中间项都消去了只剩首尾Sn=[n*(n+1)(n+2)(n+3)-0*1*2*3]/4=n*(n+1)(n+2)(n+3)/4把n=1代入S1=6=1*2*3对的

1*2*3+2*3*4+3*4*5+.+n(n+1)(n+2) =1/4【1*2*3*4-0*1*2*3】+1/4【2*3*4*5-1*2*3*4】+1/4【3*4*5*6-2*3*4*5】+.+ 1/4【n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)】 =1/4n(n+1)(n+2)(n+3)

an=n(n+1)(n+2)=(n^2+n)(n+2)=n^3+3n^2+2n sn=(1^3+2^3+3^3+……+n^3)+3(1^2+2^2+3^2+……+n^2)+2(1+2+3+……+n) =(1+2+3+……+n)^2+n(n+1)(2n+1)/2+n(n+1) =(1/4)n^2(n+1)^2+n(n+1)(2n+1)+n(n+1) =n(n+1)[(1/4)n(n+1)+2n+1+1] =n(n+1)(n^2+n+8n+8)/4 =n(n+1)(n^2+9n+8)/4

解:裂项法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

因为 n*(n+1)*(n+2)=n^3 +3n^2 +2n所以1*2*3+2*3*4+3*4*5+.+n*(n+1)*(n+2)=(1^3 +2^3 +3^3++n^3)+3(1^2 +2^2 +3^2++n^2)+2(1+2+3++n)=n^2*(n+1)^2 /4 +3*n*(n+1)*(2n+1)/6 +2*n(n+1)/2=(n^4+6n^3 +11n^2 +6n)/4

^先知道几个公式(a + b) * (a - b) = a^2 - b^2 ……(1)1 + 2 + …… + n = n(n+1)/2 …… (2)1^3 + 2^3 + …… n^3 = [n (n+1) / 2]^2 ……(3) 先用(1)把k(k+1)(k+2)看作是(k+1) * (k+1 -1) * (k+1 +1)=(k+1) * [(k+1)^2 -1]=(k+1)^3 - (k+1) 反复运用

应该是1*2+2*3+3*4++n(n+1) 吧一.n(n+1)=n^2+n 原式=(1^2+1)+(2^2+2) +(1+2+3++n) 分组求和,根据公式1^2+2^2+3^2+4^2+5^2+n^2=n(n+1)(2n+1)/6以

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