Question

设数列an的前n项和为sn，且sn=2an-2^n(n为正整数) 1.求证数列（an+1-2an）是等比数列

2，证明a1/2+a2/3+a3/4+....+an/(n+1)＜2^n

Answer 1

解：1、sn=2an-2^n
令n=1，得s1=a1=2a1-2则a1=2
得s(n+1)=2a(n+1)-2^(n+1)
两式相减得a(n+1)=2an+2^n
变形得a(n+1)/2^n=an/2^(n-1)+1
故{an/2^(n-1)}是首项为a1/1=2公差为1的等差数列
则an/2^(n-1)=n+1
则an=(n+1)*2^(n-1)
2、an/(n+1）=2^(n-1)
则a1/2+a2/3+a3/4+....+an/(n+1)=2^n-1<2^n 得证

Answer 2

1.Sn=2an-2^n，①
S<n+1>=2a<n+1>-2^(n+1),②
②-①，a<n+1>=2a<n+1>-2an-2^n,
∴a<n+1>-2an=2^n,③
∴数列{a<n+1>-2an}是等比数列。
2.由①，n=1时a1=S1=2a1-2,∴a1=2.
由③，a<n+1>/2^(n+1)-an/2^n=1/2,
∴an/2^n=a1/2+(n-1)/2=(n+1)/2,
∴an/(n+1)=2^(n-1),
∴a1/2+a2/3+a3/4+....+an/(n+1)
=1+2+2^2+……+2^(n-1)
=2^n-1＜2^n.

Answer 3

令n=1，得s1=a1=2a1-2则a1=2
得s(n+1)=2a(n+1)-2^(n+1)
两式相减a(n+1)=2an+2^n
变形得a(n+1)/2^n=an/2^(n-1)+1
an/2^(n-1)}是首项为a1/1=2公差为1的等差数列
an/2^(n-1)=n+1
an=(n+1)*2^(n-1)
an/(n+1）=2^(n-1)
则a1/2+a2/3+a3/4+....+an/(n+1)=2^n-1<2^n 得证