Question

求这道题的过程及答案

Answer 1

S2=3a1+1
a1+a2=3a1+1
a2=2a1+1，又a1+a2=4，因此a1+2a1+1=4
a1=1
S(n+1)=3Sn+1

S(n+1)+½=3(Sn+½)
[S(n+1)+½]/(Sn+½)=3，为定值。
S1+½=1+½=3/2，数列{Sn+½}是以3/2为首项，3为公比的等比数列
Sn+½=(3/2)·3ⁿ⁻¹=½·3ⁿ
Sn=½(3ⁿ-1)
n≥2时，an=Sn-S(n-1)=½(3ⁿ-1)-½(3ⁿ⁻¹-1)=3ⁿ⁻¹
n=1时，3¹⁻¹=1，a1=1同样满足表达式
数列{an}的通项公式为an=3ⁿ⁻¹
(2)
bn=2an+2log3(an)+1
=2·3ⁿ⁻¹+2log3(3ⁿ⁻¹)+1
=2·3ⁿ⁻¹+2n-2+1
=2·3ⁿ⁻¹+2n-1
Tn=2·(1+3+...+3ⁿ⁻¹)+[1+3+...+(2n-1)]
=2·1·(3ⁿ-1)/(3-1) +n²
=3ⁿ+n²-1