已知数列{an}满足a1=1,a(n+1)an=2^n,Sn是数列{an}的前n项和,S2012是多少?

题目

已知数列{an}满足a1=1,a(n+1)an=2^n,Sn是数列{an}的前n项和,S2012是多少?

答案

a2a1=2 a2=2/a1=2/1=2
a(n+1)an=2ⁿ a(n+2)a(n+1)=2^(n+1)
[a(n+2)a(n+1)]/[a(n+1)an]=a(n+2)/an=2^(n+1)/2ⁿ=2,为定值.数列的奇数项是以1为首项,2为公比的等比数列,偶数项是以2为首项,2为公比的等比数列.
S2012=(a1+a3+...+a2011)+(a2+a4+...+a2012)
=1×(2^1006 -1)/(2-1) +2×(2^1006 -1)/(2-1)
=3×2^1006 -3