已知an等于(2n-1)*(1/3)^n,求sn

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已知an等于(2n-1)*(1/3)^n,求sn
已知an等于(2n-1)*(1/3)^n,求sn

已知an等于(2n-1)*(1/3)^n,求sn
a(n)=(2n-1)×(1/3)^n

S(n)=1×(1/3)+3×(1/3)^2+…+(2n-1)×(1/3)^n
(1/3)S(n)=1×(1/3)^2+3×(1/3)^3+…+(2n-3)×(1/3)^n+(2n-1)×(1/3)^(n+1)
上式-下式,得
(2/3)S(n)=1/3+2×(1/3)^2+2×(1/3)^3+…+2×(1/3)^n-(2n-1)×(1/3)^(n+1)
=1/3-(2n-1)×(1/3)^(n+1)+2[(1/3)^2+(1/3)^3+…+(1/3)^n]
=1/3-(2n-1)×(1/3)^(n+1)+2[1+1/3+(1/3)^2+(1/3)^3+…+(1/3)^n]-2(1+1/3)
=-7/3-(2n-1)×(1/3)^(n+1)+3-(1/3)^n
=2/3-2(n+1)(1/3)^(n+1)
貌似不大对.

利用公式an=SN-S(N-1) N大于或等于2得 an=(2an-4n 1)-(2an-1 4 n=1时由SN=2an-4n 1,即a1=2a1-3得a1=3 a1 a2=2a2-7得a2=

an=(2n-1)*(1/3)^n=2n*(1/3)^n-(1/3)^n
Sn=2*[n*(1/3)^(n+1)-(n+)*(1/3)^n+1]/(1/3-1)^2/3 - (1/3)*[1-(1/3)^n]/(1-1/3)
=1-(n+1)*(1/3)^n

所以Sn=1-(n+1)*(1/3)^n