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*)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]-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={1,n=13n−1,n≥2(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]-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
=33(1−3n−1)1−3-(n-1)•32
=92•3n-9n-92,
∴S=94•3n-94-92n,
于是bn=94•3n-94-92n-n(n−3)2=94•3n-12•n2-3n-94,
∴数列{bn}的通项bn=94•3n-12•n2-3n-94.
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. | √63 | B. | 2√63 | C. | √66 | D. | √68 |
A. | -5 | B. | -25 | C. | -32 | D. | 5 |
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 |