求数列{(2n-1)*1/4的n次方}的前n项和Sn

1个回答

  • an = (2n-1)(1/4)^n

    = n(1/4)^(n-1) - (1/4)^n

    Sn =a1+a2+..+an

    = [summation(i:1->n){i(1/4)^(i-1)} ] - (1/3)(1- (1/4)^n)

    consider

    [x^(n+1)-1]/(x-1) = 1+x+x^2+...+x^n

    [(x^(n+1)-1)/(x-1)]' = 1+2x+..+nx^(n-1)

    1+2x+..+nx^(n-1) = [nx^(n+1)-(n+1)x^n+1]/(x-1)^2

    put x=1/4

    [summation(i:1->n){i(1/4)^(i-1)} ]

    =1(1/4)^0 + 1(1/4)^1+..+n(1/4)^(n-1)

    = (16/9)[n.(1/4)^(n+1)-(n+1).(1/4)^n +1 ]

    = (16/9) - (4/9)(n+2)(1/4)^n

    Sn = [summation(i:1->n){i(1/4)^(i-1)} ] - (1/3)(1- (1/4)^n)

    = (16/9) - (4/9)(n+2)(1/4)^n - (1/3)(1- (1/4)^n)

    = 13/9 - (1/9)( 4n+5)(1/4)^n .