Question

一道高中数列题~请教各路高人~~

已知数列{an}的前n项和为Sn,满足Sn=2an-2n(n∈N+)(1)求证:数列{an+2}为等比数列;(2)若数列{bn}满足bn=log2(an+2),Tn为数列{bn/(an+2)}的前n项和，求证：Tn≥1/2

Answer 1

1.答:由Sn=2An-2可得S(n-1)=2A(n-1)-2,且n≥2.
又Sn-S(n-1)=An可得2An-2A(n-1)=An....且n≥2
即可以推出An=2A(n-1),又令n=1时S1=2A1-2=A1,可知A1=2,又令n=2可得A2=4.那么An是以A2为首项,以2为公比的等比数列.
又A2=2A1,则A1也符合该数列.所以,An=A1×2^(n-1)=2^n.
2 缩放法
bn=log2(2^n+2)
Tn=bn/(an+2)=log2(2^n+2)/(an+2)=log2(2^n+2)/(2^n+2)
>=1/2

Answer 2

Sn=2an-2n(n∈N+) -----(1)
=>S(n-1)=2a(n-1)-2(n-1) (N>=2) ---(2)
(1)-(2) Sn-S(n-1)=2an-2a(n-1)+2 => an=2an-2a(n-1)+2
=> an=2a(n-1)+2 => an+2=2a(n-1)+4 =>an+2=2(a(n-1)+2)
an+2为等比数列(n>=2)，公比为2.
首项为根据Sn=2an-2n计算，令n=1, =>a1=2a1-2*1 =>a1=2

an=a1q^(n-1)=2*2^(n-1)=2^n

bn=log2(2^n+2)
Tn=bn/(an+2)=log2(2^n+2)/(an+2)=log2(2^n+2)/(2^n+2)
>=1/2