Question

两道数列题求解

Answer 1

1.
a(n+1)=4an+2
a(n+1)+2/3=4an+8/3=4(an+ 2/3)
[a(n+1) +2/3]/(an +2/3)=4，为定值
a1+2/3=2+2/3=8/3
数列{an +2/3}是以8/3为首项，4为公比的等比数列
an +2/3=(8/3)×4^(n-1)=(2/3)×4ⁿ
an=(2/3)(4ⁿ-1)
数列{an}的通项公式为an=(2/3)(4ⁿ-1)
2.
设等差数列{an}公差为d，数列为正项数列，则首项a1>0，公差d>0
S3=3a2=12
a2=4
2a1，a2，a3+1成等比数列，则
a2²=2a1·(a3+1)
a2²=2(a2-d)(a2+d+1)
a2=4代入，整理，得
d²+d-12=0
(d+4)(d-3)=0
d=-4(舍去)或d=3
an=a1+(n-1)d=a2+(n-2)d=4+3(n-2)=3n-2
数列{an}的通项公式为an=3n-2
bn=an/3ⁿ=(3n-2)/3ⁿ
Tn=b1+b2+...+bn=(3-2)/3+(3×2-2)/3²+(3×3-2)/3³+...+(3n-2)/3ⁿ
Tn/3=(3-2)/3²+(3×2-2)/3³+...+[3(n-1)-2]/3ⁿ+(3n-2)/3^(n+1)
Tn-Tn/3=(2/3)Tn
=(3-2)/3+3/3²+3/3³+...+3/3ⁿ-(3n-2)/3^(n+1)
=1/3+1/3+1/3²+...+1/3^(n-1) -(3n-2)/3^(n+1)
Tn=(3/2)[1/3+1/3+1/3²+...+1/3^(n-1) -(3n-2)/3^(n+1)]
=(1/2)[1+1+1/3+...+1/3^(n-2) -(3n-2)/3ⁿ]
=(1/2)[1+(1- 1/3^(n-1))/(1-1/3) -(3n-2)/3ⁿ]
=(5×3ⁿ-6n-5)/(4×3ⁿ)