已知等差数列{an}的前n项和为Sn,且a2=1,S11=33.(1)求{an}的通项公式（2）设bn=(1/4)^an,求证：{bn}是等比数列,并求其前n项和Tn
问题描述：
已知等差数列{an}的前n项和为Sn,且a2=1,S11=33.
(1)求{an}的通项公式
（2）设bn=(1/4)^an,求证：{bn}是等比数列,并求其前n项和Tn
答
(1)等差数列{an}S11=11a1+55d=33a2=a1+d=1d=1/2a1=1/2an=(1/2)+(1/2)(n-1)=(1/2)n(2)bn=(1/4)^anbn+1=(1/4)^an+1bn+1/bn=(1/4)^(an+1)/(1/4)^(an)=(1/4)^(an+1-an)=(1/4)^d所以：{bn}是等比数列b1=1/2Tn=(1/2)(1-(1/...