Question

4．設(shè)等比數(shù)列{a_n}的通項(xiàng)公式為a_n=2^n-1，記b_n=2（log₂a_n+1）（n∈N^*），證明：對(duì)任意的n∈N^*，不等式$\frac{_{1}+1}{_{1}}$•$\frac{_{2}+1}{_{2}}$…$\frac{_{n+1}}{_{n}}$＞$\sqrt{n+1}$成立．

Answer 1

分析 由已知結(jié)合對(duì)數(shù)的運(yùn)算性質(zhì)求得b_n=2n，得到$\frac{_{n}+1}{_{n}}=\frac{2n+1}{2n}$，代入T_n=$\frac{_{1}+1}{_{1}}$•$\frac{_{2}+1}{_{2}}$…$\frac{_{n+1}}{_{n}}$，兩邊平方后放大，再開(kāi)方得答案．

解答 證明：∵a_n=2^n-1，
b_n=2（log₂a_n+1）=$2（lo{g}_{2}{2}^{n-1}+1）=2n$，
∴$\frac{_{n}+1}{_{n}}=\frac{2n+1}{2n}$，
令T_n=$\frac{_{1}+1}{_{1}}$•$\frac{_{2}+1}{_{2}}$…$\frac{_{n+1}}{_{n}}$=$\frac{3}{2}×\frac{5}{4}×…×\frac{2n+1}{2n}$，
則${{T}_{n}}^{2}=（\frac{3}{2}）^{2}×（\frac{5}{4}）^{2}×…×（\frac{2n+1}{2n}）^{2}$$＞（\frac{3}{2}×\frac{4}{3}）（\frac{5}{4}×\frac{6}{5}）…（\frac{2n+1}{2n}×\frac{2n+2}{2n+1}）$=$\frac{2n+2}{2}=n+1$．
∴T_n=$\frac{_{1}+1}{_{1}}$•$\frac{_{2}+1}{_{2}}$…$\frac{_{n+1}}{_{n}}$$＞\sqrt{n+1}$．

點(diǎn)評(píng) 本題是數(shù)列與不等式的綜合題，考查了對(duì)數(shù)的運(yùn)算性質(zhì)，考查了放縮法證明數(shù)列不等式，是中檔題．