Question

已知数列an满足a1=1/2,2an+1-an=1.1.求an的通项公式 2.证明:求和Sn=a

Answer 1

2a(n+1) = a(n) + 1,

2a(n+1) - 2 = a(n) - 1,

a(n+1)-1 = [a(n)-1]/2,

{a(n)-1}是首项为a(1)-1=-1/2,公比为1/2的等比数列.

a(n)-1 = (-1/2)(1/2)^(n-1) = -1/2^n,

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

s(n) = n - [1/2 + 1/2^2 + ...+ 1/2^n]

= n - (1/2)[1 + 1/2 + ...+ 1/2^(n-1)]

= n - (1/2)[1 - 1/2^n]/(1-1/2)

= n - 1 + 1/2^n