1.已知等差数列{An}的前n项和为Sn,且A2=1,S11=33.
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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
(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
(1) a2=a1+d=1
s11=11a1+55d=33----->a1+5d=3
4d=2 d=0.5
a1=0.5
an=1/2n
(2)bn=(1/2)^n
bn-1*bn+1=(bn)^2
{bn} GP
Tn=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
s11=11a1+55d=33----->a1+5d=3
4d=2 d=0.5
a1=0.5
an=1/2n
(2)bn=(1/2)^n
bn-1*bn+1=(bn)^2
{bn} GP
Tn=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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