Question

数列 an 的前n项和为sn,若向量a=(Sn,1),b=(-1,2an+2^n+1),a⊥b,{bn}=an/2^n

Answer 1

(1)

a=(Sn,1),b=(-1,2an+2^(n+1))

a.b=0

-Sn+2an + 2^(n+1) =0

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

n=1

a1= -4

an = Sn -S(n-1)

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

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

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

an/2^n-a1/1=-(n-1)

an/2^n = -(n+3)

bn=an/2^n = -(n+3)

an = -(n+3).2^n

(2)

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

=-2(n+3).2^n +2^(n+1)

=-2(n+2).2^n