Question

已知数列﹛an﹜的前n项和为Sn，满足Sn=2an-2n（n∈N+）

1 求﹛an﹜的通项公式an
2 若数列﹛bn﹜满足bn=㏒2（an+2），Tn为数列﹛bn/(an+2)﹜的前n项和
求证tn≥1/2

Answer 1

a(1)=s(1)=2a(1)-2,a(1)=2.
a(n+1)=s(n+1)-s(n)=2a(n+1)-2-2a(n),
a(n+1)=2a(n)+2
a(n+1)+2=2[a(n)+2]
{a(n)+2}是首项为a(1)+2=4,公比为2的等比数列.
a(n)+2=4*2^(n-1)=2^(n+1),
a(n)=2^(n+1)-2,

b(n)=log_{2}[a(n)+2]=log_{2}[2^(n+1)]=n+1,
c(n)=b(n)/[a(n)+2]=(n+1)/2^(n+1),
t(n)=2/2^2 + 3/2^3 + 4/2^4 + ... + n/2^n + (n+1)/2^(n+1),
2t(n)= 2/2 + 3/2^2 + 4/2^3 + ... + n/2^(n-1) + (n+1)/2^n,
t(n)=2t(n)-t(n)=2/2 + 1/2^2 + 1/2^3 + ... + 1/2^n - (n+1)/2^(n+1)
=1/2+1/2 + 1/2^2 + ... +1/2^n -(n+1)/2^(n+1)
=1/2+(1/2)[1-1/2^n]/(1-1/2) - (n+1)/2^(n+1)
=1/2+1-1/2^n - (n+1)/2^(n+1)
=3/2 - (n+3)/2^(n+1),
t(n)-1/2=1 - (n+3)/2^(n+1) = [2^(n+1) - n - 3] /2^(n+1),
2^(n+1)=(1+1)^(n+1)=1^(n+1)+(n+1)1^n + ... + (n+1)*1 + 1 >=1+(n+1) + 1 = n+3
2^(n+1)>=n+3.
t(n)-1/2 = [2^(n+1)-n-3]/2^(n+1)>=0,
t(n)>=1/2

Answer 2

an=n+1