已知等差数列{an}的前n项和为Sn,且a2=1,S11=33.
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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
(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
(1)
等差数列{an}
S11=11a1+55d=33
a2=a1+d=1
d=1/2
a1=1/2
an=(1/2)+(1/2)(n-1)
=(1/2)n
(2)
bn=(1/4)^an
bn+1=(1/4)^an+1
bn+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=1/2
Tn=(1/2)(1-(1/2)^n)/(1-(1/2))=1-(1/2)^n
等差数列{an}
S11=11a1+55d=33
a2=a1+d=1
d=1/2
a1=1/2
an=(1/2)+(1/2)(n-1)
=(1/2)n
(2)
bn=(1/4)^an
bn+1=(1/4)^an+1
bn+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=1/2
Tn=(1/2)(1-(1/2)^n)/(1-(1/2))=1-(1/2)^n
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