数列｛an｝满足a1=1,前n项和为sn,sn+1=4an+2,求a2013,

Question

数列｛an｝满足a1=1,前n项和为sn,sn+1=4an+2,求a2013,

Answer 1

S(n+1)=4an+2
n=1
S2 = 4a1+2
a1+a2 =4a1+2
a2 = 3a1+2
=5
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) ]
=>{an - 2a(n-1)} 是等比数列, q=2
an - 2a(n-1) = 2^(n-2).(a2 - 2a1)
= 3.2^(n-2)
an/2^n - a(n-1)/2^(n-1) = 3/4
=>{an/2^n} 是等差数列, d=3/4
an/2^n - a2/2^2 = 3(n-2)/4
an/2^n = (3n-1)/4
an =(3n-1).2^(n-2)
a2013= (3(2013)-1).2^(2011)
=6038.2^(2011)
=3019.2^(2012)