一道数列的应用题 高中的知识应用题应用题

2个回答

  • 由,A(n+1)=2An+3得A(n+1)+3=2(An+3)

    A(n+1)+3/(An+3)=2,是等比数列,首项为a1+3=5,公比q=2,通项公式An+3=5*2^(n-1)

    An=5*2^(n-1)-3

    nAn=5n*2^(n-1)-3n

    Sn=(5*1*2^0-3*1)+(5*2*2^1-3*2)+(5*3*2^2-3*3)+(5*4*2^3-3*4)+.+[5n*2^(n-1)-3n] (1)

    2Sn=2*(5*1*2^0-3*1)+2*(5*2*2^1-3*2)+2*(5*3*2^2-3*3)+.+2*[5n*2^(n-1)-3n]

    即2Sn=(5*1*2^1-2*3*1)+(5*2*2^2-3*2*2)+(5*3*2^3-2*3*3)+...5n*2^n-2*3n] (2)

    (2)-(1)得

    Sn=-(5*1*2^0-3*1)+(-5*2-3*2)+(-5*2^2-3*3)+.+(5*2^n-3n)

    =-(5+5*2+5*2^1+...+5*2^(n-1)+3+6+9+...+3(n-1)-3n+(5*2^n-3n)

    =-5[(1-2^(n-1)]/(1-2)+[3+3(n-1)]*(n-1)/2+(5*2^n-3n)

    =5-5*2^(n-1)+(3/2)*n*(n-1)+(5*2^n-3n)

    (错位相减所得),口算,难免有误,你用笔算一下就好.