1、
S(n+1)=2Sn+3n+1
S(n+1)+3(n+1)+4=2Sn+6n+8=2(Sn+3n+4)
[S(n+1)+3(n+1)+4]/(Sn+3n+4)=2,為定值.
S1+3+4=a1+3+4=1+3+4=8
數(shù)列{Sn+3n+4}是以8為首項(xiàng),2為公比的等比數(shù)列.
Sn+3n+4=8×2^(n-1)=4×2ⁿ
Sn=4×2ⁿ-3n-4
Sn-1=4×2^(n-1)-3(n-1)-4=2×2ⁿ-3n-1
an=Sn-Sn-1=4×2ⁿ-3n-4-2×2ⁿ+3n+1=2^(n+1)-3
n=1時(shí),a1=4-3=1,同樣滿足.
bn=an+3=2^(n+1)-3+3=2^(n+1)
數(shù)列{bn}的通項(xiàng)公式為bn=2^(n+1)
2、
cn=log3(bn)=log3[2^(n+1)]=(n+1)log3(2)
(cn-1)/[(n+25)cn]
=[(n+1)log3(2)-1]/[(n+25)(n+1)log3(2)]
=log3[2^(n+1)/3]/log3(2^[(n+25)(n+1)])
2^(n+1)/3-2^[(n+25)(n+1)]
=2^(n+1)[1/3-2^(n+25)]
隨n增大,2^(n+25)單調(diào)遞增,1/3-2^(n+25)0且單調(diào)遞增
2^(n+1)[1/3-2^(n+25)]