1.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
(3)设cn=1/Tn,求c1+c2+.+c10
人气：225 ℃ 时间：2020-05-27 13:19:46
解答
(1) a2=a1+d=1s11=11a1+55d=33----->a1+5d=34d=2 d=0.5a1=0.5an=1/2n(2)bn=(1/2)^n bn-1*bn+1=(bn)^2{bn} GPTn=1-(1/2)^n(3)Cn=1/(1-(1/2)^n)=1+1/(2^n-1)S10=10+1+1/3+1/7+1/15+1/31+1/63+1/127+1/255+1/511+1/1023
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