已知数列an是等比数列,前三项为a,1/2a+1/2,1/3a+1/3.数列前n项和为Sn,求lim Sn.

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(1/2a+1/2)^2=a*(1/3a+1/3)
1/4(1/a+1)^2=1/3*a*(1/a+1)
1/4(1/a+1)=1/3*a
3(1/a+1)=4a
3(1+a)=4a^2
4a^2-3a-3=0
等比数列
[1/2(a+1)]^2=a*[1/3(a+1)]
1/4*(a+1)^2=a/3*(a+1)
等比数列的项不等于0
a+1不等于0
所以(a+1)/4=a/3
a=3
所以a1=3,a2=2,q=2/3
an=a1q^(n-1)
=3*(2/3)^(n-1)
Sn=3*[1-(2/3)^n]/(1-2/3)
=9*[1-(2/3)^n]
=9-9*(2/3)^n
lim Sn=9

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