(1)/(1+x)+(2)/(1+x^2)+(4)/(1+x^4)+(8)/(1+x^8)+…+(n)/(1+x^n)-

1个回答

  • 1/(1+x)+2/(1+x^2)+4/(1+x^4)+8/(1+x^8)+…+n/(1+x^n)-2n/(1-x^2n)

    =1/(1-x)+1/(1+x)+2/(1+x^2)+4/(1+x^4)+8/(1+x^8)+…+n/(1+x^n)-2n/(1-x^2n)-1/(1-x)

    =(1+x+1-x)/(1-x^2)+2/(1+x^2)+4/(1+x^4)+8/(1+x^8)+…+n/(1+x^n)-2n/(1-x^2n)-1/(1-x)

    =2/(1-x^2)+2/(1+x^2)+4/(1+x^4)+8/(1+x^8)+…+n/(1+x^n)-2n/(1-x^2n)-1/(1-x)

    =4/(1-x^4)+4/(1+x^4)+8/(1+x^8)+…+n/(1+x^n)-2n/(1-x^2n)-1/(1-x)

    =8/(1-x^8)+8/(1+x^8)+…+n/(1+x^n)-2n/(1-x^2n)-1/(1-x)

    .

    =2n/(1-x^2n)-2n/(1-x^2n)-1/(1-x)

    =-1/(1-x)

    =1/(x-1)