Question

已知数列an前n项和Sn=4-an-1/2^（n-2）.（1）求an+1与an的关系,(2)求an的通项

Answer 1

（1）a(n+1) = S(n+1)-Sn = an+ 1/2^(n-2) - a(n+1) - 1/2^(n-1)

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

∴2a(n+1) = an - 1/2^(n-1) ------①

（2）①式两边同时乘以2^(n-1) ,则得：

2^n a(n+1) = 2^(n-1) an -1

即数列{2^(n-1) an}是以 2^(1-1) a1 = 1为首项,-1为公差的等差数列,

故 2^(n-1) an = 1-(n-1) = 2-n

∴an的通项：

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