Question

数学第二十二题第二和第三小问

Answer 1

a2=a1+d=6,a5=a1+4d=18
解得,a1=2 ,d=4

an=4n-2
2)Tn+1/2bn=1..............(1)
T(n-1)-1/2b(n-1)=1.....(2)
(2)-(1)得:bn+1/2bn-1/2b(n-1)=0
bn/b(n-1)=1/3
即{bn}是等比数列,公比=1/3
b1=T1=2/3
bn=2/3*(1/3)^(n-1)=2*(1/3)^n
3)Cn=an*bn=(4n-2)*2*(1/3)^n=4(2n-1)*(1/3)^n
Sn=4[1/3+3*(1/3)^2+5*(1/3)^3+...+(2n-3)*(1/3)^(n-1)+(2n-1)*(1/3)^n]
1/3Sn=4[(1/3)^2+3*(1/3)^3+...+(2n-5)*(1/3)^(n-1)+(2n-1)*(1/3)^(n+1)]
2/3Sn=4/3+8[(1/3)^2+(1/3)^3+...+(1/3)^n-(2n-1)*(1/3)^(n+1)]
=4/3+4/3*[1-(1/3)^n]-(2n-1)*(1/3)^(n+1)]=8/3-[(2n+3)*(1/3)^(n+1)]
Sn=4-[(2n+3)/2*(1/3)^n]