- 试题详情及答案解析
- (本小题满分12分)已知数列{an}满足an=2an-1+2n+1(n∈N,n>1),a3=27,数列{bn}满足bn=(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.- 答案:(1)bn=+n;(2)Sn=(2n-1)×2n-n+1
- 试题分析:(1)利用{bn}成等差数列,先求出t的值,进而得到b1和公差,即可求得通项公式;(2)根据(1),可以求出{an}的通项公式,然后利用错位相减法可求出Sn.
试题解析:(1)由a3=27,得27=2a2+23+1,于是a2=9
∴9=2a1+22+1,∴a1=2
于是b1=(2+t),b2=(9+t),b3=(27+t)
∵{bn}成等差数列,故2b2=b1+b3
即2×(9+t)=(2+t)+(27+t)
解得t=1,∴b1=,b2=
bn-bn-1=1=d
∴bn=+(n-1)=+n(n∈N*)
(2)∵bn=(an+1)=+n
∴an=(n+)·2n-1=(2n-1)·2n-1-1
∴Sn=(3×20-1)+(5×21-1)+(7×22-1)+……+[(2n+1)×2n-1-1]
=3×20+5×21+7×22+……+(2n+1)×2n-1-n
2Sn= 3×21+5×22+7×23+……+(2n+1)×2n-n
作差:-Sn=3+2×2+2×22+2×23+……+2×2n-1-(2n+1)×2n+n
=1+2×-(2n+1)×2n+n
=(1-2n)×2n+n-1
∴Sn=(2n-1)×2n-n+1(n∈N*)
考点:等差数列,递推数列,通项公式,数列的前n项和,错位相减法求和