二、填空题
7.(2013·广东六校联考)设曲线y=xn+1(nN*)在点(1,1)处的切线与x轴的交点的横坐标为xn,则log2013x1+log2013x2+…+log2013x2012的值为________.
[答案] -1
[解析] 因为y′=(n+1)xn,所以在点(1,1)处的切线的斜率k=n+1,
所以=n+1,所以xn=,
所以log2013x1+log2013x2+…+log2013x2012
=log2013(x1·x2·…·x2012)
=log2013(··…·)
=log2013=-1.
8.(2014·中原名校二次联考)若{bn}为等差数列,b2=4,b4=8.数列{an}满足a1=1,bn=an+1-an(nN*),则a8=________.
[答案] 57
[解析] bn=an+1-an,a8=(a8-a7)+(a7-a6)+…+(a2-a1)+a1=b7+b6+…+b1+a1.
由{bn}为等差数列,b2=4,b4=8知bn=2n
数列{bn}的前n项和为Sn=n(n+1).
a8=S7+a1=7×(7+1)+1=57.
9.(2014·辽宁省协作校联考)若数列{an}与{bn}满足bn+1an+bnan+1=(-1)n+1,bn=,nN+,且a1=2,设数列{an}的前n项和为Sn,则S63=________.
[答案] 560
[解析] bn==,又a1=2,a2=-1,a3=4,a4=-2,a5=6,a6=-3,…,
S63=a1+a2+a3+…a63=(a1+a3+a5+…+a63)+(a2+a4+a6+…+a62)=(2+4+6+…+64)-(1+2+3+…+31)=1056-496=560.
三、解答题
10.(2014·豫东、豫北十所名校联考)已知Sn为数列{an}的前n项和,且a2+S2=31,an+1=3an-2n(nN*)
(1)求证:{an-2n}为等比数列;
(2)求数列{an}的前n项和Sn.
[解析] (1)由an+1=3an-2n可得
an+1-2n+1=3an-2n-2n+1=3an-3·2n=3(an-2n),
又a2=3a1-2,则S2=a1+a2=4a1-2,
得a2+S2=7a1-4=31,得a1=5,a1-21=3≠0,
=3,故{an-2n}为等比数列.
(2)由(1)可知an-2n=3n-1(a1-2)=3n,故an=2n+3n,
Sn=+=2n+1+-.