已知等比数列{An}中,A4=14,前10项和S10=185（1）求An（2）已知数列{Bn}的通项公式Bn=2ˆn•An,求{Bn}的前n项和Tn

Question

已知等比数列{An}中,A4=14,前10项和S10=185
（1）求An（2）已知数列{Bn}的通项公式Bn=2ˆn•An,求{Bn}的前n项和Tn

Answer 1

an :等差数列?
an=a1+(n-1)d
a4=14
a1+3d=14 (1)
S10=185
(2a1+9d)5 =185
2a1+9d =37 (2)
(2)-2(1)
3d=9
d=3
a1= 5
an = 5+3(n-1) = 3n+2
let
S =1.2^1+2.2^2+...+n.2^n (1)
2S = 1.2^2+2.2^3+...+n.2^(n+1) (2)
(2)-(1)
S= n.2^(n+1) -(2+2^2+...+2^n)
=n.2^(n+1) -2(2^n-1)
bn=2^n .an
= (3n+2).2^n
= 3(n.2^n) + 2^(n+1)
Tn =b1+b2+...+bn
=3S + 4(2^n-1)
=3n.2^(n+1) -2(2^n-1)
= 2 + (6n-2).2^n