在数列an中 S（n+1）=4an + 2 ,a1=1 设Cn=an/2的N次方 求证Cn是等差数列 求AN的通向公式及前N项和

Sn=4a(n-1)+2 S2=4a1+2a2=3a1+2=5
a(n+1)=S(n+1)-Sn=4an-4a(n-1)
a(n+1)-2an=2[an-2a(n-1)}
则{a(n+1)-2an}是公比为2的等比数列
首项=a2-2a1=5-2=3
所以a(n+1)-2an=3*2^(n-1)
a(n+1)/2^(n+1)-an/2^n=3/4
已知cn/2^n
所以c(n+1)-cn=3/4
所以{cn}是公差为3/4的等差数列
首项=c1=a1/2=1/2
所以cn=an/2^n=1/2+(3/4)(n-1)
an=(1/4)(3n-1)*2^n
Sn=(1/4)[2*2+5*2^2+8*2^3+...+(3n-1)*2^n]
(1/2)Sn=(1/4)[2+5*2+8*2^2+.+(3n-1)*2^(n-1)]
(1/2)Sn-Sn=(1/4)[2+3*2+3*2^2+...+3*2^(n-1)-(3n-1)*2^n]
-(1/2)Sn=(1/4)[1+3*(2^n-1)/(2-1)-(3n-1)*2^n]
Sn=-(1/2)[3*2^n-2-(3n-1)*2^n]
=(1/2)[(3n-4)*2^n+2]
=(1/2)(3n-4)*2^n+1