已知数列{an}满足an=2an-1+2n+1(n∈N,n>1),a3=27,数列{bn}满足bn=1/2n(an+t). (1)若数列{bn}为等差数列,求bn; (2)在(1)的条件下,求数列{an}的前n项和Sn.
已知数列{a
n}满足a
n=2a
n-1+2
n+1(n∈N,n>1),a
3=27,数列{b
n}满足b
n=
(a
n+t).
(1)若数列{b
n}为等差数列,求b
n;
(2)在(1)的条件下,求数列{a
n}的前n项和S
n.
数学人气:196 ℃时间:2020-06-11 07:04:23
优质解答
(1)由a
n=2a
n-1+2
n+1,a
3=27得,27=2a
2+2
3+1,解得a
2=9,
同理得,9=2a
1+2
2+1,解得a
1=2,
∵b
n=
(a
n+t),∴b
1=
(a
1+t)=
(2+t),
b
2=
(a
2+t)=
(9+t),b
3=
(a
3+t)=
(27+t),
∵数列{b
n}为等差数列,∴2b
2=b
1+b
3.
即
(9+t)=(2+t)+(27+t),解得t=1,
则
b1=,
b2=,即公差是1,
∴
bn=+n−1=n+;
(2)由(1)得,
bn=n+=
(a
n+1),
解得
an=(n+)•2n−1=(2n+1)•2
n-1-1,
∴S
n=(3•2
0-1)+(5•2-1)+(7•2
2-1)+…[+(2n+1)•2
n-1-1]
则S
n=3+5•2+7•2
2+…+(2n+1)•2
n-1-n ①,
2S
n=3•2+5•2
2+7•2
3+…+(2n+1)•2
n-2n ②,
①-②得,-S
n=3+2(2+2
2+2
3+…+2
n-1)-(2n+1)2
n+n
=
3+2×−(2n+1)2n+n=(1-2n)•2
n+n-1,
则S
n=(2n-1)•2
n-n+1.
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