Question

已知数列{an}是首项a1=1/4，公比q=1/4的等比数列，设(bn)+2=3log1/2an（n∈N*),数列{cn}满足cn=an*bn.

1.求数列{cn}的前n项和Sn;
2.求证：当n大于等于2时，cn+1<cn

Answer 1

1.
an=(1/4)(1/4)^(n-1)=1/4ⁿ
bn +2=3log(1/2)an=3log(1/2)(1/4ⁿ)=3log(1/2)[(1/2)^(2n)]=3(2n)=6n
bn=6n-2
cn=anbn=(6n-2)/4ⁿ=6n/4ⁿ -2/4ⁿ
Sn=c1+c2+...+cn=6(1/4 +2/4²+3/4³+...+n/4ⁿ) -2(1/4+1/4²+...+1/4ⁿ)
令Cn=1/4+2/4²+3/4³+...+n/4ⁿ
则Cn/4=1/4²+2/4³+...+(n-1)/4ⁿ+n/4^(n+1)
Cn-Cn/4=(3/4)Cn=1/4+1/4²+...+1/4ⁿ -n/4^(n+1)
=(1/4)[1-(1/4)ⁿ]/(1-1/4) -n/4^(n+1)
=(1/3)(1-1/4ⁿ) -(n/4)/4ⁿ
Cn=(4/9)(1-1/4ⁿ)-(n/3)/4ⁿ
Sn=6[(4/9)(1-1/4ⁿ)-(n/3)/4ⁿ]-2(1/4)(1-1/4ⁿ)/(1-1/4)
=(8/3)(1-1/4ⁿ) -(2n)/4ⁿ -(2/3)(1-1/4ⁿ)
=8/3 -[(8/3)+2n-(2/3)]/4ⁿ -2/3
=2 -[(8+6n-2)/3]/4ⁿ
=2- 2(n+1)/4ⁿ
2.
证：
c(n+1)-cn=[6(n+1)-2]/4^(n+1) -(6n -2)/4ⁿ
=[6(n+1)-2-4(6n-2)]/4^(n+1)
=(6n+6-2-24n+8)/4^(n+1)
=(-18n+12)/4^(n+1)
=-6(3n-2)/4^(n+1)
n≥1 3n-2>0 4^(n+1)>0 6>0，因此6(3n-2)/4^(n+1)>0
-6(3n-2)/4^(n+1)<0
c(n+1)-cn<0
c(n+1)<cn，不等式成立。