已知数列{an}满足a1=2,an=2an-1+2(n∈N*,且n≥2)若数列{bn}满足bn=log2(an+2)设T

证：
n≥2时,
an=2a(n-1)+2
an+2=2a(n-1)+4=2[a(n-1)+2]
(an +2)/[a(n-1)+2]=2,为定值.
a1+2=2+2=4
数列{an +2}是以4为首项,2为公比的等比数列.
an +2=4×2^(n-1)=2^(n+1)
bn=log2(an +2)=log2[2^(n+1)]=n+1
bn/(an +2)=(n+1)/2^(n+1)
Tn=b1/(a1+2)+b2/(a2+2)+...+bn/(an+2)
=2/2²+3/2³+4/2⁴+...+(n+1)/2^(n+1)
Tn/2=2/2³+3/2⁴+...+n/2^(n+1)+(n+1)/2^(n+2)
Tn-Tn/2=Tn/2=1/2 +1/2³+...+1/2^(n+1) -(n+1)/2^(n+2)
Tn=1+1/2²+1/2³+...+1/2ⁿ- (n+1)/2^(n+1)
=1/2 +(1/2+1/2²+1/2³+...+1/2ⁿ) -(n+1)/2^(n+1)
=1/2 +(1/2)(1-1/2ⁿ)/(1-1/2) -(n+1)/2^(n+1)
=3/2 -1/2ⁿ -(n+1)/2^(n+1)