一道数列让我晕…

a(n) = aq^(n-1), 0<q<1.
a(1)+a(2)+...+a(n) = a[1-q^n]/(1-q)
9 = a/(1-q), a = 9(1-q),
a(n) = 9q^(n-1) - 9q^n = 9(1-q)q^(n-1),
[a(n)]^2 = 81(1-q)^2q^[2(n-1)]
[a(1)]^2+[a(2)]^2+...+[a(n)]^2=81(1-q)^2[1-q^(2n)]/(1-q^2)
81/5 = 81(1-q)^2/(1-q^2)=81(1-q)/(1+q),1+q=5(1-q),
q=2/3.a=9(1-q)=3.

a(n) = 3(2/3)^(n-1),n=1,2,...
T(k,m) = a(k) + (2k-1)(m-1) = 3(2/3)^(k-1) + (2k-1)(m-1),k=1,2,...,n, m=1,2,...
T(2,m) = a(2) + 3(m-1) = 3(2/3) + 3(m-1)= 2 + 3(m-1).
T(2,1)+T(2,2)+...+T(2,10)=2*10 + 3*9*10/2 = 155.

b(i) = T(k,i) = 3(2/3)^(k-1) + (2k-1)(i-1),i=1,2,...
S(n) = b(1)+b(2)+...+b(n)
= 3n(2/3)^(k-1) + (2k-1)(n-1)n/2

（1）{(an)^2}可以看作是首项为(a1)^2，公比为q^2的等比数列
用等比数列和公式
{an}各项和9=a1(1-q^n)/(1-q)=Sn
{(an)^2}各项和81/5=[(a1)^2][1-(q^2)^n]/(1-q^2)=Kn
解方程组，(Sn)^2/Kn,解得q=2/3
代入Sn中得a1=3
所以an=a1*[q^(n-1)]=3*[(2/3)^(n-1)]

(2)a2=3*（2/3)=2
T^2的前10项和为S=[a2+a2+9(2*a2-1)]*10/