Question

19．已知數(shù)列{a_n}滿足：a₁=1，2a_n+1a_n+3a_n+1+a_n+2=0．求證：{$\frac{1}{{a}_{n}+1}$}是等差數(shù)列．

Answer 1

分析 根據(jù)等差數(shù)列的定義進(jìn)行整理證明即可．

解答 證明：∵2a_n+1a_n+3a_n+1+a_n+2=0．
∴2a_n+1a_n=-3a_n+1-a_n-2．
則$\frac{1}{{a}_{n+1}+1}$-$\frac{1}{{a}_{n}+1}$=$\frac{{a}_{n}+1-{a}_{n+1}-1}{（{a}_{n+1}+1）（{a}_{n}+1）}$=$\frac{{a}_{n}-{a}_{n+1}}{{a}_{n+1}{a}_{n}+{a}_{n}+{a}_{n+1}+1}$=$\frac{2（{a}_{n}-{a}_{n+1}）}{2{a}_{n+1}{a}_{n}+2{a}_{n}+2{a}_{n+1}+2}$
=$\frac{2（{a}_{n}-{a}_{n+1}）}{-3{a}_{n+1}-{a}_{n}-2+2{a}_{n}+2{a}_{n+1}+2}$=$\frac{2（{a}_{n}-{a}_{n+1}）}{{a}_{n}-{a}_{n+1}}=2$為常數(shù)，
故：{$\frac{1}{{a}_{n}+1}$}是公差d=2的等差數(shù)列．

點(diǎn)評 本題主要考查等差數(shù)列的判斷，根據(jù)數(shù)列的遞推關(guān)系是解決本題的關(guān)鍵．