求和1/2+2*(1/2)^2+3*(1/2)^3…+n*(1/2)^n

2个回答

  • 设:

    S=1/2+2×(1/2)²+3×(1/2)³+…+n×(1/2)^n

    (1/2)S===(1/2)²+2×(1/2)³+…+(n-1)×(1/2)^n+n×(1/2)^(n+1)

    两式相减,得:

    (1/2)S=(1/2)+(1/2)²+(1/2)³+…+(1/2)^n-n×(1/2)^(n+1)

    (1/2)S=1-2×(1/2)^(n+1)-n×(1/2)^(n+1)

    (1/2)S=1-(n+2)×(1/2)^(n+1)

    则:

    S=2-(n+2)×(1/2)^n