题目内容
13.已知数列{an}的首项a1=1,前n项和为Sn,且Sn+1-3Sn-2n-4=0(n∈N*)(1)求数列{an}的通项公式;
(2)设函数f(x)=anx+an-1x2+an-2x3+…+a1xn,f′(x)是函数f(x)的导函数,令bn=f′(1),求数列{bn}的通项公式.
分析 (1)当n≥2时,通过Sn+1-3Sn-2n-4=0,Sn-3Sn-1-2n-2=0相减,得an+1=3an+2,结合a1=1,可得a2=8,进而可得结论;
(2)通过求导可知bn=an+2an-1+…+na1=[3n+2•3n-1+3•3n-2+…+(n-1)•32]-$\frac{n(n-3)}{2}$,利用错位相减法计算出3n+2•3n-1+3•3n-2+…+(n-1)•32,进而可得结论.
解答 解:(1)当n≥2时,由Sn+1-3Sn-2n-4=0,得Sn-3Sn-1-2n-2=0,
两式相减,得an+1=3an+2,
∴an+1+1=3(an+1)(n≥2)成立,
又已知a1=1,a2=8,而a2+1≠3(a1+1),
∴数列{an+1}从第二项起是以9为首项,3为公比的等比数列,
∴数列{an}的通项an=$\left\{\begin{array}{l}{1,}&{n=1}\\{{3}^{n}-1,}&{n≥2}\end{array}\right.$(n∈N*);
(2)∵函数f(x)=anx+an-1x2+an-2x3+…+a1xn,
∴bn=f′(1)=an+2an-1+…+na1
=(3n-1)+2(3n-1-1)+3(3n-2-1)+…+(n-1)(32-1)+n
=[3n+2•3n-1+3•3n-2+…+(n-1)•32]-[1+2+3+…+(n-1)]+n
=[3n+2•3n-1+3•3n-2+…+(n-1)•32]-[1+2+3+…+(n-1)+n]+2n
=[3n+2•3n-1+3•3n-2+…+(n-1)•32]-$\frac{n(n-3)}{2}$
令S=3n+2•3n-1+3•3n-2+…+(n-1)•32,
则3S=3n+1+2•3n+3•3n-1+4•3n-2+…+(n-1)•33,
两式相减得:2S=3n+1+3n+3n-1+3n-2+…+33-(n-1)•32
=$\frac{{3}^{3}(1-{3}^{n-1})}{1-3}$-(n-1)•32
=$\frac{9}{2}$•3n-9n-$\frac{9}{2}$,
∴S=$\frac{9}{4}$•3n-$\frac{9}{4}$-$\frac{9}{2}$n,
于是bn=$\frac{9}{4}$•3n-$\frac{9}{4}$-$\frac{9}{2}$n-$\frac{n(n-3)}{2}$=$\frac{9}{4}$•3n-$\frac{1}{2}$•n2-3n-$\frac{9}{4}$,
∴数列{bn}的通项bn=$\frac{9}{4}$•3n-$\frac{1}{2}$•n2-3n-$\frac{9}{4}$.
点评 本题考查求数列的通项,考查运算求解能力,注意解题方法的积累,属于中档题.
A. | $\frac{{\sqrt{6}}}{3}$ | B. | $\frac{{2\sqrt{6}}}{3}$ | C. | $\frac{{\sqrt{6}}}{6}$ | D. | $\frac{{\sqrt{6}}}{8}$ |
A. | -5 | B. | -$\frac{2}{5}$ | C. | -$\frac{3}{2}$ | D. | 5 |
①对于命题P:存在x∈R,使得x2+x-1<0,则﹁P:任意x∈R,均有x2+x-1>0
②命题“若x=y,则sinx=siny”的逆否命题为真命题
③“m=-1”是“直线l1:mx+(2m-1)y+1=0与直线l2:3x+my+3=0垂直”的充要条件.
A. | 3个 | B. | 2个 | C. | 1个 | D. | 0个 |
x | 1 | 2 | 3 | 4 | 5 | 6 |
y | 3 | 7 | 5 | 9 | 6 | 1 |