已知数列｛an｝的首项是a1=1,前n项和为Sn,且S(n+1)=2Sn+3n+1

题目

已知数列｛an｝的首项是a1=1,前n项和为Sn,且S(n+1)=2Sn+3n+1
1）设bn=an+3,求bn通项公式2)设cn=log3bn,若存在常数k,使不等式k>=(cn-1)/[(n+25)cn]恒成立,求k的最小值

答案

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,为定值.
S1+3+4=a1+3+4=1+3+4=8
数列{Sn+3n+4}是以8为首项,2为公比的等比数列.
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时,a1=4-3=1,同样满足.
bn=an+3=2^(n+1)-3+3=2^(n+1)
数列{bn}的通项公式为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)]
随n增大,2^(n+25)单调递增,1/3-2^(n+25)0且单调递增
2^(n+1)[1/3-2^(n+25)]