设数列{an}的前n项和Sn与an的关系是Sn=-an+1-1/2^n(1)求an

Question

设数列{an}的前n项和Sn与an的关系是Sn=-an+1-1/2^n (1)求an

Answer 1

a(1) = s(1) = -a(1) + 1 - 1/2,2a(1) = 1/2,a(1) = 1/4.
s(n) = -a(n) + 1 - 1/2^n,
s(n+1) = -a(n+1) + 1 - 1/2^(n+1),
a(n+1) = s(n+1)-s(n) = -a(n+1) - 1/2^(n+1) + a(n) + 1/2^n,
2a(n+1) = a(n) + 1/2^(n+1),
2^(n+1)a(n) = 2^na(n) + 1/2,
{2^na(n)}是首项为2a(1)=1/2,公差为1/2的等差数列.
2^na(n) = 1/2 + (n-1)/2 = n/2,
a(n) = n/2^(n+1).