哪位大神来设数列｛an｝的前n项和为Sn,n∈N*，已知a1=1，a2=3/2，a3=5/4，且当n≥2时，4S(n+2)+5Sn=8S（n+1）+S（n-1）。（1）求a4的值（2）证明：｛a（n+1）-1/2an｝为等比数列（3）求数列｛an｝的通项

Question

哪位大神来
设数列｛an｝的前n项和为Sn,n∈N*，已知a1=1，a2=3/2，a3=5/4，且当n≥2时，4S(n+2)+5Sn=8
S（n+1）+S（n-1）。
（1）求a4的值
（2）证明：｛a（n+1）-1/2an｝为等比数列
（3）求数列｛an｝的通项公式
第一小问不用做，本人会

Answer 1

4S(n+2)+5Sn=8S(n+1) +S(n-1)
4[S(n+2) -S(n+1)] = 4[S(n+1)- Sn] - [Sn -S(n-1)]
4a(n+2) = 4a(n+1) - an
a(n+2) = a(n+1) - (1/4)an
a(n+2) -(1/2)a(n+1) = (1/2)[ a(n+1) - (1/2)an ]
=>{a(n+1) - (1/2)an} 是等比数列, q=1/2
a(n+1) - (1/2)an = (1/2)^(n-2) .(a3 - (1/2)a2)
=(1/2)^(n-2) .(5/4 - (1/2)(3/2))
=(1/2)^(n-1)
2^(n+1)a(n+1) -2^n. an = 2
=>{ 2^n. an } 是等差数列, d=2
2^n. an - 2^3. a3 = 2(n-3)
2^n. an - 10 = 2(n-3)
2^n. an = 2(n+2)
an = (n+2)/2^(n-1)
ie
an =1 ; n=1
=3/2 ; n=2
=(n+2)/2^(n-1) ; n>=3
a4 = (4+2)/2^3
=6/8
=3/4