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  • 在數(shù)列{an}中,a1=1,an+1=(1+1/n)an+(n+1)/(2^n) (1) 設(shè)bn=an/n,求數(shù)列{bn}的通項(xiàng)公式

    在數(shù)列{an}中,a1=1,an+1=(1+1/n)an+(n+1)/(2^n) (1) 設(shè)bn=an/n,求數(shù)列{bn}的通項(xiàng)公式
    在數(shù)列{an}中,a1=1,an+1=(1+1/n)an+(n+1)/(2^n)
    (1)設(shè)bn=an/n,求數(shù)列{bn}的通項(xiàng)公式
    (2)求數(shù)列{an}的前n項(xiàng)和sn
    數(shù)學(xué)人氣:896 ℃時(shí)間:2019-08-21 23:19:57
    優(yōu)質(zhì)解答
    (1)
    a(n+1)=(1+1/n)an+(n+1)/(2^n)
    a(n+1)/(n+1) = (1/n)an + 1/(2^n)
    a(n+1)/(n+1) - (1/n)an = 1/(2^n)
    an/n - a(n-1)/(n-1) = 1/2^(n-1)
    an/n - a1/1 = 1/2^(n-1) +1/2^(n-2)+..+ 1/2^1
    = 1- 1/2^(n-1)
    an/n = 2- 1/2^(n-1) = bn
    (2)
    an/n = 2- 1/2^(n-1)
    an = 2n - n(1/2)^(n-1)
    consider
    1+x+x^2+...+x^n = (x^(n+1) -1) /(x-1)
    1+2x+..+n.x^(n-1)
    =[(x^(n+1) -1) /(x-1)]'
    = { nx^(n+1) - (n+1)x^n + 1 } / (x-1)^2
    put x= 1/2
    1.(1/2)^0 + 2(1/2)^1+..+n(1/2)^(n-1)
    = 4(n.(1/2)^(n+1) - (n+1)(1/2)^n + 1 )
    an = 2n - n(1/2)^(n-1)
    Sn = n(n+1) - 4(n.(1/2)^(n+1) - (n+1)(1/2)^n + 1 )
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