Question

已知数列｛an｝的前n项和Sn=-an-（1／2）的n－1次方再加2,令bn＝2的n次方乘an,求证bn等差,

Answer 1

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

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

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

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

两边同乘以2的n次方

得到2^(n+1)·a(n+1) - 2^n·an = b(n+1) - bn = 1

S1 = a1 = -a1 +1 得 a1 = 0.5

b1 = 2a1 =1

bn = n = 2^n·an 得 an = 2^(-n)