Question

高中数学数列题，求高手解答，手写优先采纳，拜托写整齐。拜托各位尽快啊！！

Answer 1

Answer 2

a1+2a3+3a3+...+(n-1)a(n-1)+nan=(n-1)Sn+3n
a1+2a3+3a3+...+(n-1)a(n-1)=(n-2)S(n-1)+3(n-1)
两式相减得:

nan=(n-1)Sn-(n-2)S(n-1)+3
n[Sn-S(n-1)]=(n-1)Sn-(n-2)S(n-1)+3
nSn-nS(n-1)=(n-1)Sn-(n-2)S(n-1)+3
Sn-2S(n-1)-3=0
Sn=2S(n-1)+3
Sn+3=2[S(n-1)+3]
(Sn+3)/[S(n-1)+3]=2
所以,{Sn+3}是等比数列
2) a1=S1=3
Sn+3=(S1+3)*2^(n-1)=(3+3)*2^(n-1)=3*2^n
bn=(12n-3)/(Sn+3)=(12n-3)/3*2^n=(4n-1)*(1/2)^n
Tn=3*(1/2)+7*(1/2)^2+11*(1/2)^2...+(4n-1)*(1/2)^n
1/2*Tn=3*(1/2)^2+7*(1/2)^3+...+(4n-5)*(1/2)^n-4n-1)*(1/2)^(n+1)
两式相减得:
1/2*Tn=3*(1/2)+4[(1/脊迅2)^2+(1/2)^3+...+(1/2)^n]-4n-1)*(1/2)^(n+1)
1/2*Tn=-1/2+4[1/2+(1/2)^2+(1/2)^3+...+(1/2)^n]-(4n-1)*(1/2)^(n+1)
1/2*Tn=-1/2+4{[1/2(1-(1/2)^n]/(1-1/2)}-(4n-1)*(1/2)^(n+1)
1/2*Tn=-1/2+4[(1-(1/2)^n]-(4n-1)*(1/2)^(n+1)
1/2*Tn=7/2-4(1/2)^n-(4n-1)/2*(1/2)^n
1/2*Tn=7/2-4*(1/2)^n-(4n-1)/2*(1/2)^n
1/巧伏2*Tn=7/2-(4n+7)/孝野携2*(1/2)^n
Tn=7-[(4n+7)*(1/2)^n]<7
所以,Tn<7

Answer 3

可以拍清楚点吗