d=(a4-a2)/2=4 則an=5+4(n-2)=4n-3
由Tn+bn=3 得T(n+1)+b(n+1)=3
兩式相減得b(n+1)=1/2bn
令n=1得T1+b1=2b1=3則b1=3/2
故{bn}是首項(xiàng)為3/2公比為1/2的等比數(shù)列.
則bn=3/2^n
則Cn=(4n-3)*3/2^n
故C(n+1)=(4n+1)*3/2^(n+1)
則C(n+1)/Cn=(4n+1)/(8n-6)
令(4n+1)/(8n-6)>=1解得n<=7/4
故當(dāng)n=1時(shí),C(n+1)>Cn
當(dāng)n>=2時(shí),C(n+1)