已知数列{an}为等比数列,前三项为a,1/2a+1/2,1/3a+1/3,且Sn=a1+a2+a3+…+an,则Tn=
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已知数列{an}为等比数列,前三项为a,1/2a+1/2,1/3a+1/3,且Sn=a1+a2+a3+…+an,则Tn=a1^2+a2^2+…+an^2等于?
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
所以bn=an^2,则b1=9,
bn/b(n-1)=[an/a(n-1)]^2=q^2=4/9
所以Tn=9*[1-(4/9)^n]/(1-4/9)=81/5-(81/5)*(4/9)^n
[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
所以bn=an^2,则b1=9,
bn/b(n-1)=[an/a(n-1)]^2=q^2=4/9
所以Tn=9*[1-(4/9)^n]/(1-4/9)=81/5-(81/5)*(4/9)^n
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