已知S=1/2-2/4+3/8-4/16+...+(-1)^(k-1)(k/2^k)+...+(2005/2^2005)

1个回答

  • S=1/2-2/4+3/8-4/16+...+(-1)^(k-1)(k/2^k)+...+(2005/2^2005)-(2006/2^2006)

    (1/2)*S=1/4-2/8+3/16-4/32+...+(-1)^(k-1)[k/2^(k+1)]+...+(2005/2^2006)-(2006/2^2007)

    两式相加,得

    (3/2)S=1/2-1/4+1/8-1/16+…………+1/2^2005-1/2^2006+1/2^2007

    =(1/2+1/8+1/32+…………+1/2^2005+1/2^2007)-(1/4+1/16+…………+1/2^2004+1/2^2006)

    =(1/2)[1-(1/4)^1004]/(1-1/4)-(1/4)[1-(1/4)^1003]/(1-1/4)

    =(2/3)*[1-(1/4)^1004]-(1/3)*[1-(1/4)^1003]

    =1/3+(1/6)*(1/4)^1003

    ∴S=2/9+(1/9)*(1/4)^1003

    结果还可以写成S=2/9+1/(9*4^1003)或者写成S=2/9+1/(9*2^2006)

    容易看出:

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