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，

还要求an

Answer 1

解：
S1 = a1 = -a1 - (1/2)^0 + 2,a1 = 1/2
Sn = -an - (1/2)^(n-1) + 2
S(n-1)=-a(n-1)-(1/2)^(n-2)+2
相减：
an = a(n-1)-an + (1/2)^(n-1)
∴an = a(n-1)/2 + (1/2)^n
上式是包含a(n-1)的递推式
an =(1/2)^n + [a(n-2)/2+(1/2)^(n-1)]/2
=2*(1/2)^n + a(n-2)/4
=3*(1/2)^n + a(n-3)/饥行或带高8
=.........
=(n-1)*(1/烂伍2)^n + a1/2^(n-1)
=n/(2^n)
bn = 2^n * an = 2^n * n/(2^n) = n
b(n-1) = n-1
即：bn - b(n-1) = 1
所以bn是首项为1公差为1的等差数列

Answer 2

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)
两首睁边同乘睁困以2的n次方
得到者早岁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)

追问

an=n/(2^n)

追答

呃 是我疏忽了 不好意思