问题
解答题
| （文）已知等差数列{an}的首项a1=0且公差d≠0，bn=2^an（n∈N*），Sn是数列{bn}的前n项和． （1）求Sn； （2）设Tn=
|
答案
（文）（1）an=（n-1）d，bn=2^an=2（n-1）d（4分）
Sn=b1+b2+b3+…+bn=20+2d+22d+…+2（n-1）d
由d≠0得2d≠1，∴Sn=
| 1-(2d)n |
| 1-2d |
（2）Tn=
| Sn |
| bn |
| ||
| 2(n-1)d |
| 1-2nd |
| 2(n-1)d-2nd |
∴
| lim |
| n→∞ |
| lim |
| n→∞ |
| 1- 2nd |
| 2nd-d-2nd |
| lim |
| n→∞ |
| 1-(2d)n |
| (2d)n-1-(2d)n |
=
| lim |
| n→∞ |
| ||
|
| 2d |
| 2d-1 |
G
G