已知Sn为数列{an}的前项n和,且Sn=2an+n^2-3n-2(n∈N*),令bn=an-2n(n∈N*)（1）求证：数列{bn}为等比数例；（2）令Cn=1/(bn+1),记Tn=C1C2+2C2C3+2^2C3C4+……+2^（n-1)CnCn+1,试比较 Tn与1/6的大小.

已知Sn为数列{an}的前项n和,且Sn=2an+n^2-3n-2(n∈N*),令bn=an-2n(n∈N*)（1）求证：数列{bn}为等比数例；（2）令Cn=1/(bn+1),记Tn=C1C2+2C2C3+2^2C3C4+……+2^（n-1)CnCn+1,试比较 Tn与1/6的大小.
已知Sn为数列{an}的前项n和,且Sn=2an+n^2-3n-2(n∈N*),令bn=an-2n(n∈N*)
（1）求证：数列{bn}为等比数例；
（2）令Cn=1/(bn+1),记Tn=C1C2+2C2C3+2^2C3C4+……+2^（n-1)CnCn+1,试比较 Tn与1/6的大小.

已知Sn为数列{an}的前项n和,且Sn=2an+n^2-3n-2(n∈N*),令bn=an-2n(n∈N*)（1）求证：数列{bn}为等比数例；（2）令Cn=1/(bn+1),记Tn=C1C2+2C2C3+2^2C3C4+……+2^（n-1)CnCn+1,试比较 Tn与1/6的大小.
(1) Sn=2an+n^2-3n-2 S(n-1)=2a(n-1)+(n-1)^2-3(n-1)-2
Sn-S(n-1)=2an-2a(n-1)+n^2-(n-1)^2-3n+3(n-1)=2an-2a(n-1)+2n-4=an
an=2a(n-1)-2n+4 an-2n=2a(n-1)-4n+4=2a(n-1)-4(n-1)=2[a(n-1)-2(n-1)]
(an-2n)/[a(n-1)-2(n-1)]=2 所以数列{bn}是公比为2的等比数列.
(2) n=1时 有S1=a1=2a1+1-3-2=2a1-4 a1=4 b1=2 bn=2*2^(n-1)=2^n>0
Cn=1/(2^n+1)>0 C(n+1)=1/[2^(n+1)+1]
CnC(n+1)=1/{(2^n+1)[2^(n+1)+1]}=(1/2^n){1/(2^n+1)-1/[2^(n+1)+1]}
2^(n-1)CnC(n+1)=(1/2){1/(2^n+1)-1/[2^(n+1)+1]}
Tn=(1/2){1/3-1/5+1/5-1/9+.+1/(2^n+1)-1/[2^(n+1)+1]}=(1/2){1/3-1/[2^(n+1)+1]}
=1/6-(1/2){1/[2^(n+1)+1]}

当n=1时 ： a1 = S1 = 2a1 + 1^2 - 3*1 - 2 => a1 = 4
当n>1时 ： an = Sn - S(n-1) = (2an + n^2 - 3n - 2) - [2a(n-1) + (n-1)^2 - 3(n-1) - 2]
= 2an - 2a(n-1) + n^2 - (n-1)...

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当n=1时 ： a1 = S1 = 2a1 + 1^2 - 3*1 - 2 => a1 = 4
当n>1时 ： an = Sn - S(n-1) = (2an + n^2 - 3n - 2) - [2a(n-1) + (n-1)^2 - 3(n-1) - 2]
= 2an - 2a(n-1) + n^2 - (n-1)^2 - 3n + 3(n-1) -2 + 2
= 2an - 2a(n-1) + 2n - 4
=> an = 2an - 2a(n-1) + 2n - 4
an = 2a(n-1) - 2n + 4
bn=an-2n => an = bn + 2n, b1 = a1 - 2 = 2. 当n>1时,代入上式：
bn + 2n = 2[b(n-1) + 2(n-1)] - 2n + 4
bn = 2b(n-1) , 又b1 = 2
=> bn为等比数列且bn = 2^n

(2) Cn=1/(bn+1) => Cn = 1/(1+ 2^n)
2^(n-1)CnC(n+1) = 2^(n-1) / [(1+2^n)(1+2^(n+1))] = 1/2 * [1/(1+2^n) - 1/[1+2^(n+1)]]
=> 2Tn = 1/(1+2^1) - 1/(1+2^2) + 1/(1+2^2) - 1/(1+2^3) + ... + [1/(1+2^n) - 1/[1+2^(n+1)]]
=> 2Tn = 1/3 - 1/[1+2^(n+1)]
=> Tn = 1/6 - 1/2[1+2^(n+1)] < 1/6

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当n=1时 ： a1 = S1 = 2a1 + 1^2 - 3*1 - 2 => a1 = 4
当n>1时 ： an = Sn - S(n-1) = (2an + n^2 - 3n - 2) - [2a(n-1) + (n-1)^2 - 3(n-1) - 2]
= 2an - 2a(n-1) + n^2 - (n-1)...

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当n=1时 ： a1 = S1 = 2a1 + 1^2 - 3*1 - 2 => a1 = 4
当n>1时 ： an = Sn - S(n-1) = (2an + n^2 - 3n - 2) - [2a(n-1) + (n-1)^2 - 3(n-1) - 2]
= 2an - 2a(n-1) + n^2 - (n-1)^2 - 3n + 3(n-1) -2 + 2
= 2an - 2a(n-1) + 2n - 4
=> an = 2an - 2a(n-1) + 2n - 4
an = 2a(n-1) - 2n + 4
bn=an-2n => an = bn + 2n, b1 = a1 - 2 = 2. 当n>1时,代入上式：
bn + 2n = 2[b(n-1) + 2(n-1)] - 2n + 4
bn = 2b(n-1) , 又b1 = 2
=> bn为等比数列且bn = 2^n

(2) Cn=1/(bn+1) => Cn = 1/(1+ 2^n)
2^(n-1)CnC(n+1) = 2^(n-1) / [(1+2^n)(1+2^(n+1))] = 1/2 * [1/(1+2^n) - 1/[1+2^(n+1)]]
=> 2Tn = 1/(1+2^1) - 1/(1+2^2) + 1/(1+2^2) - 1/(1+2^3) + ... + [1/(1+2^n) - 1/[1+2^(n+1)]]
=> 2Tn = 1/3 - 1/[1+2^(n+1)]
=> Tn = 1/6 - 1/2[1+2^(n+1)] < 1/6

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