bài này giải tnao

a, Ta sẽ chứng minh $\displaystyle u_{n} >1,\ \forall n\in N^{*}$ Thật vậy với $\displaystyle n=1\Rightarrow u_{1} =3 >1$ Ta sẽ giả sử $\displaystyle u_{n} >1$ $\displaystyle u_{n+1} -1=\frac{3u_{n} +1}{u_{n} +3} -1=\frac{2u_{n} -2}{u_{n} +3} =\frac{2( u_{n} -1)}{u_{n} +3} >0$ Vậy theo nguyên lý quy nạp $\displaystyle \Rightarrow u_{n} >1,\ \forall n\in N^{*}$ Xét $\displaystyle u_{n+1} -u_{n}$ $\displaystyle \begin{array}{{>{\displaystyle}l}} =\frac{3u_{n} +1}{u_{n} +3} -u_{n} =\frac{1-u_{n}^{2}}{u_{n} +3}\\ u_{n} >1\\ \Rightarrow u_{n}^{2} >1\\ \Rightarrow 1-u_{n}^{2} < 0\\ \Rightarrow u_{n+1} -u_{n} < 0\\ \Rightarrow u_{n+1} < u_{n} \end{array}$ vậy đây là dãy giảm b, Xét $\displaystyle u_{n+1} -1=\frac{2( u_{n} -1)}{u_{n} +3}$ $\displaystyle \begin{array}{{>{\displaystyle}l}} u_{n+1} +1=\frac{3u_{n} +1}{u_{n} +3} +1=\frac{4u_{n} +4}{u_{n} +3} =\frac{4( u_{n} +1)}{u_{n} +3}\\ \Rightarrow \frac{u_{n+1} +1}{u_{n+1} -1} =2\frac{u_{n} +1}{u_{n} -1} \end{array}$ do đó $\displaystyle \begin{array}{{>{\displaystyle}l}} \frac{u_{n+1} +1}{u_{n+1} -1} =2\frac{u_{n} +1}{u_{n} -1} =2^{2}\frac{u_{n-1} +1}{u_{n-1} -1} =...=2^{n}\frac{u_{1} +1}{u_{1} -1} =2^{n+1}\\ \Rightarrow 1+\frac{2}{u_{n+1} -1} =2^{n+1} \Rightarrow \frac{1}{u_{n+1} -1} =2^{n} -\frac{1}{2}\\ \Rightarrow S=\left( 2^{0} -\frac{1}{2}\right) +\left( 2^{1} -\frac{1}{2}\right) +...+\left( 2^{99} -\frac{1}{2}\right) =2^{100} -1-50=2^{100} -51 \end{array}$