Question

19．已知数列{a_n}满足：a₁=1，2a_n+1a_n+3a_n+1+a_n+2=0．求证：{$\frac{1}{{a}_{n}+1}$}是等差数列．

Answer 1

分析 根据等差数列的定义进行整理证明即可．

解答 证明：∵2a_n+1a_n+3a_n+1+a_n+2=0．
∴2a_n+1a_n=-3a_n+1-a_n-2．
则$\frac{1}{{a}_{n+1}+1}$-$\frac{1}{{a}_{n}+1}$=$\frac{{a}_{n}+1-{a}_{n+1}-1}{（{a}_{n+1}+1）（{a}_{n}+1）}$=$\frac{{a}_{n}-{a}_{n+1}}{{a}_{n+1}{a}_{n}+{a}_{n}+{a}_{n+1}+1}$=$\frac{2（{a}_{n}-{a}_{n+1}）}{2{a}_{n+1}{a}_{n}+2{a}_{n}+2{a}_{n+1}+2}$
=$\frac{2（{a}_{n}-{a}_{n+1}）}{-3{a}_{n+1}-{a}_{n}-2+2{a}_{n}+2{a}_{n+1}+2}$=$\frac{2（{a}_{n}-{a}_{n+1}）}{{a}_{n}-{a}_{n+1}}=2$为常数，
故：{$\frac{1}{{a}_{n}+1}$}是公差d=2的等差数列．

点评 本题主要考查等差数列的判断，根据数列的递推关系是解决本题的关键．