(1)S(2n) = 2n(a1+a2n)/2 = n(a1+a2n)
Sn = n(a1+an)/2
S(2n)/Sn = 2(a1+a2n)/(a1+an) = 4
a1+a2n =2a1+2an
a2n = an + nd,其中d為等差公比；
∴an +nd = a1 + 2an
an = nd - a1
而：an = a1 + (n-1)d
對比：d = 2
所以：an = 2n-1
Sn = n^2
(2)bn = an*2^(n-1) = (2n-1) *2^(n-1)
Tn = 2^0 + 3*2^1 + 5 * 2^2 + 7*2^3+.+(2n-1) *2^(n-1)
2Tn =2 + 3**2^2+5 * 2^3+.+(2n-3)*2^(n-1)+(2n-1) *2^n
兩式相減：
-Tn = 2^0 + 2*2 + 2*2^2 +.+2^2^(n-1)-(2n-1) *2^n
Tn = -1+(2n-1) *2^n - 4(2^n -1)為什么S(2n) = 2n(a1+a2n)/2 = n(a1+a2n)？