Question

设数列an的前n项和为Sn，a1=1,Sn+1=4an+2(n∈N+)

Answer 1

a(1)=1,
1+a(2)=a(1)+a(2)=s(2)=4a(1)+2=4+2=6, a(2)=5.
s(n+1)=4a(n)+2,
s(n+2)=4a(n+1)+2,
a(n+2)=s(n+2)-s(n+1)=4a(n+1)-4a(n),
a(n+2)-2a(n+1)=2a(n+1)-4a(n)=2[a(n+1)-2a(n)],
{a(n+1)-2a(n)}是首项为a(2)-2a(1)=5-2=3,公比为2的等比数列.
a(n+1)-2a(n)=3*2^(n-1).
a(n+1)=2a(n)+3*2^(n-1).
a(n+1)/2^n=a(n)/2^(n-1) + 3/2,
{a(n)/2^(n-1)}是首项为a(1)=1,公差为(3/2)的等差数列.
a(n)/2^(n-1) =1+(3/2)(n-1)=(3n-1)/2.
a(n) = (3n-1)2^(n-2),