Question

设数列an的前n项和为sn，满足sn=2an-2的n次方，令bn=an/2的n次方 求证bn为等差数列

求数列an的通项公式

Answer 1

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

Answer 2

sn=2an-2^n
s(n-1)=2a(n-1)-2^(n-1)
an=2an-2^n-2a(n-1)+2^(n-1)
2a(n-1)-2^(n-1)=an-2^n
[a(n-1)/2^(n-1)-1/2]*2^n=(an/2^n-1)*2^n
an/2^n-1=a(n-1)/2^(n-1)-1/2
an/2^n-a(n-1)/2^(n-1)=1/2
bn=an/2^n
bn-b(n-1)=1/2
数列bn为等差数列。
a1=2a1-2 a1=2 b1=1 d=1/2
bn=b1+(n-1)d=1+n/2-1/2=(n+1)/2
an/2^n=(n+1)/2
an=(n+1)*2^(n-1)