设an,an+k(k为常数)均为等比数列,若a1=2,Sn是an的前n项和,且K不等于0,则S(3n-1)-bn=

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  • 设 等比数列an的公比为q

    an+k(k为常数)均为等比数列

    所以:

    Bn=an+k(k为常数) 的公比为:(2q^n+k)/(2q^(n-1) +k)

    (2q^n+k)/(2q^(n-1) +k) 为不等于0的常数

    即:

    (2q^n+k)/(2q^(n-1) +k)

    =[q( 2q^(n-1) +k ) +k-kq ] /(2q^(n-1) +k)

    =q+ k(1-q)/ (2q^(n-1) +k) 为不等于0的常数

    即k(1-q)/ (2q^(n-1) +k)得值与n的值无关

    令n=1 k(1-q)/ (2q^(n-1) +k) =1-q (k不等于0)

    令n=2 k(1-q)/ (2q^(n-1) +k) = k(1-q)/ (2q +k)

    则令:1-q = k(1-q)/ (2q +k) (q不等于0)

    得出:q=1

    把q=1代入式k(1-q)/ (2q^(n-1) +k) = 0 其值与n的值无关,满足条件

    即:bn的公比为:(2q^n+k)/(2q^(n-1) +k)

    =q+ k(1-q)/ (2q^(n-1) +k)

    =q

    =1

    即bn=a1+k=2+k

    S(3n-1)-bn=2(3n-1)--(2+k)=6n-k