Question

高二数列题在线等急！

Answer 1

(1)
a1=1
S(n+1) = 4an +2
a1+a2+....+a(n+1)= 4an +2 (1)
a1+a2+....+an= 4a(n-1) +2 (2)
(1)-(2)
a(n+1)= 4an -4a(n-1)
a(n+1)-2an = 2[an - 2a(n-1) ]
=> {a(n+1)-2an } 是等比数列， q=2
(2)
a1=1
S2 = 4a1 +2
a1+a2 =4a1 +2
a2 = 3a1 +2 = 5
a(n+1)-2an = 2^(n-1) .(a2-2a1)
=2^(n-1) .(5-2)
= 3.2^(n-1)
a(n+1)/2^(n+1) - an/2^n = 3/4
=> {an/2^n} 是等差数列, d=3/4
(3)
an/2^n - a2/2^2 = (3/4)(n-2)
an/2^n - a2/2^2 = (3/4)(n-2)
an/2^n - 5/4 = (3/4)(n-2)
an/2^n = (3/4)n -(1/4)
an = [(3/4)n -(1/4)] .2^n
=(1/4)( 3n-1).2^n

追问

最后一步的前n项和怎么求?

我知道啦谢谢

Answer 2

追问

谢谢