Question

数列{an}满足a1=a2=1,an=1-(a1+.+a(n-2))/4(n>=3),则an=

Answer 1

an=1-[a1+a2+...+a(n-2)]/4 (1)
a(n+1)=1-[a1+a2+...+a(n-1)]/4 (2)
(2)-(1)
a(n+1)-an=-a(n-1)/4
4a(n+1)-4an=-a(n-1)
4a(n+1)-2an=2an-a(n-1)
[2a(n+1)-an]/[2an-a(n-1)]=1/2,为定值.
2a2-a1=2-1=1
数列{2a(n+1)-an}是以1为首项,1/2为公比的等比数列.
2a(n+1)-an=(1/2)^(n-1)
等式两边同乘以2ⁿ
a(n+1)×2^(n+1)-an×2ⁿ=2,为定值.
a1×2=1×2=2
数列{an×2ⁿ}是以2为首项,2为公差的等差数列.
an×2ⁿ=2+2(n-1)=2n
an=2n/2ⁿ
n=1时,a1=2/2=1；n=2时,a2=4/4=1,同样满足.
数列{an}的通项公式为an=2n/2ⁿ.