设an公差不为0的等差数列.(1)前n项和为Sn,Sn=110,a1.a2.a4为等比数列.求an通项公式.

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设公差d
a2 = a1 + d
a4 = a1 +3d
a10 = a1 + 9d
S10 = (a1 +a10)*10/2 = 5(2a1 + 9d) = 110
2a1 + 9d = 22
a1.a2.a4为等比数列
a2 * a2 = a1 * a4
(a1 + d)^2 = a1(a1 + 3d)
d^2 - a1*d = 0
d = a1
( d = 0 舍去)
2*a1 + 9*a1 = 22
a1 = 2
d = 2
an = a1 + (n-1)d = 2 + (n-1)*2 = 2n
bn = n * 2^(2n) = n*4^n
Tn = b1 + b2 + …… + bn
= 4 + 2*4^2 + 3*4^3 + …… +(n-1)*4^(n-1) + n*4^n
Tn/4 = 1 + 2*4 + 3*4^2 + …… + (n-1)*4^(n-2) + n*4^(n-1)
Tn - Tn/4 = -1 + 4*(1-2) + 4^2 *(2-3) + 4^3*(3-4) + …… + 4^(n-1)*[(n-1) -n] + n*4^n
= n*4^n - 1 -[4 + 4^2 + 4^3 + …… + 4^(n-1)]
= n*4^n - 1 - 4*[4^(n-1) -1]/(4-1)
= n*4^n -1 - 4^n /3 + 4/3
= (n - 1/3)*4^n + 1/3
Tn = (4/3) * [(n -1/3)*4^n + 1/3]
结果检验：
b1 = 4
b2 = 2*4^2 = 32
b3 = 3*4^3 = 192
T1 = 4
T2 = 36
T3 = 228
按照推出的 Tn 计算
Tn = (4/3) * [(n -1/3)*4^n + 1/3]
T1 = (4/3) * [(1 - 1/3)*4 + 1/3] = (4/3)*(8/3 + 1/3) = =4
T2 = (4/3) * [(2 - 1/3)*4^2 + 1/3] = (4/3)*(80/3 + 1/3) = 36
T3 = (4/3) * [(3 - 1/3)*4^3 + 1/3] = (4/3)*(512/3 + 1/3) = 228
检验成立
Tn = (4/3) * [(n -1/3)*4^n + 1/3]
= （4/9） * [(3n -1)*4^n + 1]