an=(n-1/2)*2的n次方,求Sn,并证明Sn/2的n次方>2n-3.

1个回答

  • an = (n-1/2).2^n

    = 2(n.2^(n-1)) - 2^(n-1)

    consider

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

    1+2x+..+nx^(n-1)

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

    ={nx^(n+1)-(n+1)x^n +1 } /(x-1)^2

    put x=2

    1.2^0+2.2^1+...+n.2^(n-1) = n2^(n+1)-(n+1)2^n +1

    an = 2(n.2^(n-1)) - 2^(n-1)

    Sn = a1+a2+..+an

    = 2{n2^(n+1)-(n+1)2^n +1} - (2^n-1)

    = (2n-3).2^n +3

    sn/2^n = (2n-3) + 3/2^n

    > 2n-3