1、
把(Sn,S(n+1))代入y=(n+1)x/n + n+1,得：
S(n+1)=(n+1)Sn/n + n+1
两边同时除以n+1,得
S(n+1)/(n+1)=Sn/n + 1
∴数列{Sn/n}是以S1/1=3为首项,1为公差的等差数列.
2、
Sn/n=3 + (n-1)×1 = n + 2
Sn=n(n+2)=n²+2n
S(n-1)=(n-1)(n+1)=n²-1
∴an=Sn-S(n-1)=n²+2n - (n²-1)=2n+1 ,n≥2
把a1=3代入也满足.
∴an=2n+1
∴bn=an·2^an = (2n+1)·2^(2n+1)
Tn=3×2^3 + 5×2^5 + 7×2^7 + …… + (2n+1)×2^(2n+1)
4Tn= 3×2^5 + 5×2^7 + …… + (2n-1)×2^(2n+1) + (2n+1)×2^(2n+3)
两式相减,得：
3Tn= (2n+1)×2^(2n+3) - 3×2^3 - [2^6 + 2^8 + 2^(2n+2)]
= (2n+1)×2^(2n+3) - 3×2^3 - [4^(n+2) - 2^6]/3
= (2n+1)×2^(2n+3) - 3×2^3 - [2^(2n+4) - 2^6]/3
∴Tn=[(2n+1)/3]×2^(2n+3) - 2^3 - [2^(2n+4) - 2^6]/9
=[(2n+1)/3]×2^(2n+3) - 2^3 - [2^(2n+4)]/9 + (2^6)/9
∴Cn=Tn / 2^(2n+3)
= (2n+1)/3 - 1/2^(2n) - 2/9 + 2^3/(9×2^2n)
= (2n+1)/3 - 1/2^(2n) - 2/9 + 8/(9×2^2n)
= (6n+1)/9 - 1/2^(9×2n)
= (1/9) [ (6n+1) - 1/4^n ]
分开求和：
数列Dn=6n+1 前n项和为Rn=n(D1+Dn)/2=(7+6n+1)n/2=3n²+4n
数列En=1/4^n 前n项和为Qn=E1(1-q^n)/(1-q)=(1/4)[1-(1/4^n)]/(3/4) = (4^n - 1)/(3×4^n)
∴C1 + C2 + C3 + ……+Cn
＝(1/9) (Rn - Qn)
＝(1/9) [3n²+4n - (4^n - 1)/(3×4^n) ]
＝(1/27) [9n²+12n - (4^n - 1)/4^n ] (提取1/3)
＝(1/27) (9n²+12n - 1 + 1/4^n ) (把后面分式拆开)
＞(1/27) (9n²+12n - 1 ) (∵1/4^n＞0)
≥(1/27) (9+12 - 1 ) (∵上式明显单调递增)
＝20/27