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材料;1x2=1/3x(1x2x3%0x1x2) 2x3=1/3x(2x3x4%1x2x...

【1】1x2+2x3+3x4+...+10x11 =10×11×12/3 =440 【2】1x2+2x3+3x4+...+nx[n+1] =n(n+1)(n+2)/3 【3】请类比以上形式,尝试连续三个字然数的乘积的规律并计算 1x2x3+2x3x4+3x4x5+...+7x8x9 =7×8×9×10/4 =1260

根据以上规律: 3X4=1/3(3X4X5-2X3X4) nX(n+1)=1/3[nX(n+1)X(n+2)-(n-1)XnX(n+1)] 1X2+2X3+3X4+、、、、、、+nX(n+1)=1/3[1X2X3-0X1X2+2X3X4-1X2X3+3X4X5-2X3X4+...+nX(n+1)X(n+2)-(n-1)XnX(n+1)]=1/3[nX(n+1)X(n+2)-0X1X2]=n(n+1)(n+2)/3

一般的,有: (n-1)n(n+1) =n^3-n {n^3}求和公式:Sn=[n(n+1)/2]^2 {n}求和公式:Sn=n(n+1)/2 1x2x3+2x3x4+3x4x5+....+7x8x9 =2^3-2+3^3-3+...+8^3-8 =(2^3+3^3+...+8^3)-(2+3+...+8) =[(8*9/2)^2-1]-8*9/2+1 =1260

一般的,有: (n-1)n(n+1) =n^3-n {n^3}求和公式:Sn=[n(n+1)/2]^2 {n}求和公式:Sn=n(n+1)/2 1x2x3+2x3x4+3x4x5+....+7x8x9 =2^3-2+3^3-3+...+8^3-8 =(2^3+3^3+...+8^3)-(2+3+...+8) =[(8*9/2)^2-1]-8*9/2+1 =1260

3*(1x2+2x3+3x4++99x100)=3*1/3*(1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+99x100x101-98x99x100)=99x100x101选C

3*(1x2+2x3+3x4+...+99x100) =3*1/3*(1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+99x100x101-98x99x100) =99x100x101 选 C

(1)(n-1)n(n+1)-n³ =(n²-1)n-n³ =n³-n-n³ =-n; 所以(n-1)n(n+1)-n³=-n (2)带入第一问,当n=1000时 999*1000*1001=-1000

1X2+2X3+。。。nX(n+1)=1/3XnX(n+1)X(n+2)

c语言 main() {int a,b,c,max,sum; sum=0;a=1;b=2;c=3; scanf("%d",&max); for(;max

D=-153、D1=153、D2=153、D3=0、D4=-153 ∴x1=D1/D=-1 x2=D2/D=-1 x3=D3/D=0 x4=D4/D=1

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