数列{an},a1=1,an=2-2Sn,求an,若bn=n*an,求{bn}的前n项和Tn
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数列{an},a1=1,an=2-2Sn,求an,若bn=n*an,求{bn}的前n项和Tn
因为an=2-2Sn……(1)
所以a(n-1)=2-2S(n-1)……(2)
(1)-(2)得:
an- a(n-1)= -2(Sn-S(n-1))
即an- a(n-1)= -2an推出于
an=(1/3)a(n-1)
所以是以1为首项1/3为公比的等比数列
所以an=(1/3)^(n-1)
则bn=n*(1/3)^(n-1)
Tn=(1/3)^0+2(1/3)^1+3(1/3)^2+……+n(1/3)^(n-1) …… (3)
(1/3)*Tn=(1/3)^1+2(1/3)^2+3(1/3)^3+……+(n-1) (1/3)^(n-1)+n(1/3)^n…(4)
(3)-(4)得
(2/3)Tn=1+(1/3)^1+(1/3)^2+(1/3)^3+……+(1/3)^(n-1)- n(1/3)^n
=(2/3)(1-(1/3)^n)- n(1/3)^n
所以Tn=1-(1/3)^n- (3/2)n(1/3)^n
所以a(n-1)=2-2S(n-1)……(2)
(1)-(2)得:
an- a(n-1)= -2(Sn-S(n-1))
即an- a(n-1)= -2an推出于
an=(1/3)a(n-1)
所以是以1为首项1/3为公比的等比数列
所以an=(1/3)^(n-1)
则bn=n*(1/3)^(n-1)
Tn=(1/3)^0+2(1/3)^1+3(1/3)^2+……+n(1/3)^(n-1) …… (3)
(1/3)*Tn=(1/3)^1+2(1/3)^2+3(1/3)^3+……+(n-1) (1/3)^(n-1)+n(1/3)^n…(4)
(3)-(4)得
(2/3)Tn=1+(1/3)^1+(1/3)^2+(1/3)^3+……+(1/3)^(n-1)- n(1/3)^n
=(2/3)(1-(1/3)^n)- n(1/3)^n
所以Tn=1-(1/3)^n- (3/2)n(1/3)^n
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