Question

数列{an}中,Sn=4a(n-1) +1(n≥2) 且a1=1,若bn=a(n+1)-2an.求an的通向公式

Answer 1

S(n）=4a(n-1)+1

S(n+1)=4an+1

两者相减,得

S(n+1)-S(n)=a(n+1)=4[an-a(n-1)]

bn=a（n+1）-2anb(n-1)=an-2a(n-1)

bn=a(n+1)-2an=4[an-a(n-1)]-2an=2an-4a(n-1)=2*b(n-1)

{bn}等比；公比为2的,首项a1=1,s2=4a1+2；a2=5,

b1=a2-2a1=5-2×1=3

故bn=3*2^(n-1)

bn=a(n+1)-2an=3*2^(n-1)

a(n+1)/2^(n+1)-an/2^n=3/4

{an/2^n}等差,公差3/4,首项1/2

an/2^n=1/2+(n-1)*3/4=(3n-1)/4

an=(3n-1)2^n/4