1×2=1/3 (1×2×3-0×1×2) 2×3=1/3 (2×3×4-1×2×3) 3×4=1/3 (3×4×5-2×3×4) 由以上三个等式相加,可得1×2+2×3+3×4=(1/3)×3×4×5=20 读完以上材料,请你计算(1)1×2+2×3+3×4+…+n×(n+1)=_ (2)1×2+2×3+3×4+…+10×11=_

来源：学生作业帮助网编辑：作业帮时间：2024/05/05 11:50:42

1×2=1/3 (1×2×3-0×1×2) 2×3=1/3 (2×3×4-1×2×3) 3×4=1/3 (3×4×5-2×3×4) 由以上三个等式相加,可得1×2+2×3+3×4=(1/3)×3×4×5=20 读完以上材料,请你计算(1)1×2+2×3+3×4+…+n×(n+1)=_____ (2)1×2+2×3+3×4+…+10×11=_____
1×2=1/3 (1×2×3-0×1×2) 2×3=1/3 (2×3×4-1×2×3) 3×4=1/3 (3×4×5-2×3×4) 由以上三个等式相加,可得1×2+2×3+3×4=(1/3)×3×4×5=20 读完以上材料,请你计算(1)1×2+2×3+3×4+…+n×(n+1)=_____ (2)1×2+2×3+3×4+…+10×11=_____(写出过程) (3)1×2×3+2×3×4+3×4×5+…+7×8×9=______

1×2=1/3 (1×2×3-0×1×2) 2×3=1/3 (2×3×4-1×2×3) 3×4=1/3 (3×4×5-2×3×4) 由以上三个等式相加,可得1×2+2×3+3×4=(1/3)×3×4×5=20 读完以上材料,请你计算(1)1×2+2×3+3×4+…+n×(n+1)=_____ (2)1×2+2×3+3×4+…+10×11=_____
(1)1×2+2×3+3×4+…+n×(n+1)= (1/3)n(n+1)(n+2)
(2)1×2+2×3+3×4+…+10×11= 1/3 (12×3-0×1×2) +1/3 (2×3×4-1×2×3)+…………+1/3(10x11x12-9x8x7)=1/3（10x11x12-0x1x2)=440
(3) .1260

(1)(1/3)*(n+1)*(n+2)
(2)(1/3)*10*11*12=440
(3)规律(1/4)*n*(n+1)*(n+2)*(n+3) 因此答案为(1/4)*7*8*9*10=1260

(1)1×2+2×3+3×4+…+n×(n+1)=_(n+1)(n+2)/3____ (2)1×2+2×3+3×4+…+10×11=(11+1)11_/3=44____(写出过程) (3)1×2×3+2×3×4+3×4×5+…+7×8×9=__180____

3(2y-1)-2(1-y)=0 !1 1/2x^2+1/3x-1=0 1/2x^2+1/3x-1=0 1/[(1+0)*(2+0)]+1/[(1+1)*(2+1)]+1/[(1+2)*(2+2)]+1/[(1+3)*(2+3)]+```+1/[(1+2010)*(2+2010)]解释一下为什么1/(n+1)(n+2)=1/(n+1)-1/(n+2) 解矩阵方程ax=x+bA=3 2 3 1 0 0 -1 2 2 B=2 1 3 1 -1 1 3 1 1 求值D=|1 2 3 4||1 0 1 2||3 -1 -1 0||1 2 0 -5| 5+2+0+1+3+1+4=2+5+0+1+3+1+4 求矩阵A=(1,1,2,2,1;0,2,1,5,-1;2,0,3,-1,3;1,1,0,4,-1)的秩题:矩阵的秩 .[1 1 2 2 1] A =[0 2 1 5 -1][2 0 3 -1 3][1 1 0 4 -1] ,哪行减哪哪行... 行列式 1 2 3 -1 1 -1 0 20 1 0 1 0 0 -1 2 81^1/2-(1/8)^-1+3^0=? 1/3(x+1)^2-1=0 A=矩阵1 0 1,-1 2 0 ,0 4 1B=矩阵2 1 2,-3 1 1,1 0 1 AX=B (0-1)(1-2)(2-3).(100-101)=? 2+3^(x-1)-9^(x-1/2)=0 3(2y-1)-2(1-y)=0 矩阵列变换对矩阵作初等变换 1 1 -1 1 0 0 1 0 0 1/3 1/6 2/3（A,E)=0 2 2 0 1 0 = .=0 1 1 0 1/2 0 = 1 -1 0 0 0 1 0 0 1 -1/3 1/3 1/3 c语言中b[3][3]={0,1,2,0,1,2,0,1,