Question

12．已知數(shù)列{a_n}滿足a₁=1，a_n+1=2a_n+1．
（1）證明數(shù)列{a_n+1}為等比數(shù)列；
（2）若數(shù)列{b_n}滿足b₁=a₁，$\frac{b_n}{a_n}=\frac{1}{a_1}+\frac{1}{a_2}+…+\frac{1}{{{a_{n-1}}}}（n≥2，n∈{N^*}）$．
①求b_n+1a_n-（b_n+1）a_n+1的值；
②求證：$（1+{b_1}）（1+{b_2}）•…•（1+{b_n}）＜\frac{10}{3}{b_1}•{b_2}•…•{b_n}（n∈{N^*}）$．

Answer 1

分析 （1）由a_n+1=2a_n+1得a_n+1+1=2（a_n+1），可得數(shù)列{a_n+1}是以2為首項，公比為2的等比數(shù)列．
（2）①由已知可得$\frac{_{n+1}}{{a}_{n+1}}=\frac{_{n}}{{a}_{n}}+\frac{1}{{a}_{n}}$，即可得b_n+1a_n-（b_n+1）a_n+1得值； ②由①知$\frac{_{n}+1}{_{n+1}}=\frac{{a}_{n}}{{a}_{n+1}}，（n≥2）$，b₁=a₁=1，b₂=a₂=3．
$\frac{_{1}+1}{_{1}}•\frac{_{2}+1}{_{2}}•\frac{_{3}+1}{_{3}}…\frac{_{n}+1}{_{n}}$=$\frac{1}{_{1}}•\frac{_{1}+1}{_{2}}•\frac{_{2}+1}{_{3}}…\frac{_{n}+1}{_{n+1}}•_{n+1}$=$\frac{1}{_{1}}•\frac{_{1}+1}{_{2}}•\frac{{a}_{2}}{{a}_{3}}•\frac{{a}_{3}}{{a}_{4}}…\frac{{a}_{n}}{{a}_{n+1}}•_{n+1}$=2$\frac{_{n+1}}{{a}_{n+1}}=2（\frac{1}{{a}_{1}}+\frac{1}{{a}_{2}}+…+\frac{1}{{a}_{n}}）$ 即可證得$（1+{b_1}）（1+{b_2}）•…•（1+{b_n}）＜\frac{10}{3}{b_1}•{b_2}•…•{b_n}（n∈{N^*}）$．

解答 解：（1）由a_n+1=2a_n+1得a_n+1+1=2（a_n+1）．
⇒$\frac{{a}_{n+1}+1}{{a}_{n}+1}=2$，
∴數(shù)列{a_n+1}是以2為首項，公比為2的等比數(shù)列．
（2）由（1）得數(shù)列{a_n+1}是以2為首項，公比為2的等比數(shù)列．
∴a_n+1=2•2^n-1=2ⁿ，⇒${a}_{n}={2}^{n}-1$
①，$\frac{b_n}{a_n}=\frac{1}{a_1}+\frac{1}{a_2}+…+\frac{1}{{{a_{n-1}}}}（n≥2，n∈{N^*}）$…①．
$\frac{_{n+1}}{{a}_{n+1}}\frac{1}{{a}_{1}}+\frac{1}{{a}_{2}}+..+\frac{1}{{a}_{n-1}}+\frac{1}{{a}_{n}}$…③
②-①得$\frac{_{n+1}}{{a}_{n+1}}=\frac{_{n}}{{a}_{n}}+\frac{1}{{a}_{n}}$⇒b_n+1a_n-（b_n+1）a_n+1=0（n≥2）
當(dāng)n=1時，b₁=a₁=1，b₂=a₂=3，∴b₂a₁-a₂（b₁+1）=-3．
②由①知$\frac{_{n}+1}{_{n+1}}=\frac{{a}_{n}}{{a}_{n+1}}，（n≥2）$，b₁=a₁=1，b₂=a₂=3．
∴$（\frac{1}{_{1}}+1）（\frac{1}{_{2}+1}）…（\frac{1}{_{n}}+1）$=$\frac{_{1}+1}{_{1}}•\frac{_{2}+1}{_{2}}•\frac{_{3}+1}{_{3}}…\frac{_{n}+1}{_{n}}$=$\frac{1}{_{1}}•\frac{_{1}+1}{_{2}}•\frac{_{2}+1}{_{3}}…\frac{_{n}+1}{_{n+1}}•_{n+1}$
=$\frac{1}{_{1}}•\frac{_{1}+1}{_{2}}•\frac{{a}_{2}}{{a}_{3}}•\frac{{a}_{3}}{{a}_{4}}…\frac{{a}_{n}}{{a}_{n+1}}•_{n+1}$=2$\frac{_{n+1}}{{a}_{n+1}}=2（\frac{1}{{a}_{1}}+\frac{1}{{a}_{2}}+…+\frac{1}{{a}_{n}}）$
∵k≥2時，$\frac{1}{{2}^{k}-1}＜2（\frac{1}{{2}^{k}-1}-\frac{1}{{2}^{k+1}-1}）$．
∴$\frac{1}{{a}_{1}}+\frac{1}{{a}_{2}}+…+\frac{1}{{a}_{n}}＜1+\frac{1}{2}+2$[（$\frac{1}{{2}^{2}-1}-\frac{1}{{2}^{3}-1}$）+…+（$\frac{1}{{2}^{n}-1}-\frac{1}{{2}^{n+1}-1}$）]
=1+2（$\frac{1}{3}-\frac{1}{{2}^{n+1}-1}$）$＜\frac{5}{3}$
∴$（\frac{1}{_{1}}+1）（\frac{1}{_{2}+1}）…（\frac{1}{_{n}}+1）$═2$\frac{_{n+1}}{{a}_{n+1}}=2（\frac{1}{{a}_{1}}+\frac{1}{{a}_{2}}+…+\frac{1}{{a}_{n}}）$$＜\frac{10}{3}$
∴$（1+{b_1}）（1+{b_2}）•…•（1+{b_n}）＜\frac{10}{3}{b_1}•{b_2}•…•{b_n}（n∈{N^*}）$．

點評 本題考查了利用數(shù)列遞推式求通項的方法，考查了數(shù)列與不等式，考查了計算能力，屬于中檔題．