已知x=(-1)的n次方-(-1)n+1的次方/2,求x+2x的平方+3x的三次方+.+99x的99次方+100x的10

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  • 设S=x+2x^2+3x^3+4x^4+······+100x^100,则:

    xS=x^2+2x^3+3x^4+4x^5+······+99x^100+100x^101.

    ∴S-xS=x+x^2+x^3+x^4+······+x^100-100x^101,

    ∴(1-x)S=x(1-x^100)/(1-x)-100^101,

    ∴S=x(1-x^100)/(1-x)^2-100x^101/(1-x).

    一、当n为偶数时,(-1)^n=1、(-1)^(n+1)=-1,∴此时x=2.

    ∴x+2x^2+3x^3+4x^4+······+100x^100

    =x(1-x^100)/(1-x)^2-100x^101/(1-x)

    =2(1-2^100)/(1-2)^2-100×2^101/(1-2)

    =2(1-2^100)+100×2^101

    =2-2^101+100×2^101

    =2+99×2^101.

    二、当n为奇数时,(-1)^n=-1、(-1)^(n+1)=1,∴此时x=-2.

    ∴x+2x^2+3x^3+4x^4+······+100x^100

    =x(1-x^100)/(1-x)^2-100x^101/(1-x)

    =(-2)[1-(-2)^100]/(1+2)^2-100×(-2)^101/(1+2)

    =-2(1-2^100)/9+100×2^101/3

    =(300×2^101+2^101-2)/9

    =(301×2^101-2)/9