题目内容
(2012•长春模拟)已知数列{an}满足a1=1,an+1=2an+1(n∈N*).
(1)求证:数列{an+1}是等比数列,并写出数列{an}的通项公式;
(2)若数列{bn}满足4b1-1•4b2-1•4b3-1…4bn-1=(an+1)n,求数列{bn}的前n项和Sn.
(1)求证:数列{an+1}是等比数列,并写出数列{an}的通项公式;
(2)若数列{bn}满足4b1-1•4b2-1•4b3-1…4bn-1=(an+1)n,求数列{bn}的前n项和Sn.
分析:(1)由题意可得 an+1+1=2(an+1),数列{an+1}是以2为公比、以2为首项的等比数列,求得an+1=2n,从而求得{an}的通项公式.
(2)由题意可得 4b1+b2+…+bn-n=(2n)n=(2)n2,即 2(b1+b2+…+bn)-2n=n2,由此求得数列{bn}的前n项和Sn.
(2)由题意可得 4b1+b2+…+bn-n=(2n)n=(2)n2,即 2(b1+b2+…+bn)-2n=n2,由此求得数列{bn}的前n项和Sn.
解答:解:(1)证明:∵an+1=2an+1(n∈N*),∴an+1+1=2(an+1).
又 a1=1,a1+1≠0,∴
=2,
∴数列{an+1}是以2为公比、以2为首项的等比数列,
∴an+1=2n,即an =2n-1.
(2)∵4b1-1•4b2-1•4b3-1…4bn-1=(an+1)n,
]∴4b1+b2+…+bn-n=(2n)n=(2)n2,
∴2(b1+b2+…+bn)-2n=n2,
∴b1+b2+…+bn=
+ n.
又 a1=1,a1+1≠0,∴
an+1+1 |
an+1 |
∴数列{an+1}是以2为公比、以2为首项的等比数列,
∴an+1=2n,即an =2n-1.
(2)∵4b1-1•4b2-1•4b3-1…4bn-1=(an+1)n,
]∴4b1+b2+…+bn-n=(2n)n=(2)n2,
∴2(b1+b2+…+bn)-2n=n2,
∴b1+b2+…+bn=
n2 |
2 |
点评:本题主要考查等比关系的确定,指数幂的运算性质,属于中档题.
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