已知数列{an}\{a_n\}{an}满足a1=1,an+1=3an+1.a_1=1,a_{n+1}=3a_n+1.a1=1,an+1=3an+1.
(1)证明：{an+12}\{a_n+\dfrac{1}{2}\}{an+21}是等比数列，并求{an}\{a_n\}{an}的通项公式；
(2)证明：1a1+1a2+⋯+1an<32.\dfrac{1}{a_1}+\dfrac{1}{a_2}+\cdots+\dfrac{1}{a_n}<\dfrac{3}{2}.a11+a21+⋯+an1<23.
[解析]
(1)由题意有an+1+12=3(an+12)a_{n+1}+\dfrac{1}{2}=3(a_n+\dfrac{1}{2})an+1+21=3(an+21) 所以 {an+12}\{a_n+\dfrac{1}{2}\}{an+21} 是以 32\dfrac{3}{2}23 为首项， 333 为公比的等比数列；解得 an=3n−12.a_n=\dfrac{3^n-1}{2}.an=23n−1.
(2)因为当 n>1n>1n>1时23n−1<13n−1\dfrac{2}{3^n-1}<\dfrac{1}{3^{n-1}}3n−12<3n−11则
1a1+1a2+⋯+1an\dfrac{1}{a_1}+\dfrac{1}{a_2}+\cdots+\dfrac{1}{a_n}\quad\quad\quad\quada11+a21+⋯+an1=23−1+232−1+⋯+23n−1=\dfrac{2}{3-1}+\dfrac{2}{3^2-1}+\cdots+\dfrac{2}{3^n-1}=3−12+32−12+⋯+3n−12<1+13+132+⋯+13n−1<1+\dfrac{1}{3}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{3^{n-1}}\quad\quad\quad<1+31+321+⋯+3n−11=32−12⋅3n−1<32=\dfrac{3}{2}-\dfrac{1}{2\cdot3^{n-1}}<\dfrac{3}{2}\quad\quad\quad\quad\quad\quad=23−2⋅3n−11<23当 n=1n=1n=1 时，显然成立；得证.
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