a: \(\dfrac{u_n}{u_{n-1}}=\dfrac{3^n}{2^{n+1}}:\dfrac{3^{n-1}}{2^n}\)

\(=\dfrac{3^n}{3^{n-1}}\cdot\dfrac{2^n}{2^{n+1}}=\dfrac{3}{2}>1\)

=>(un) là dãy tăng

c: ĐKXĐ: n>=1

\(u_n=\sqrt{n}-\sqrt{n-1}\)

\(=\dfrac{1}{\sqrt{n}+\sqrt{n-1}}\)

\(\dfrac{u_n}{u_{n-1}}=\dfrac{1}{\sqrt{n}+\sqrt{n-1}}:\dfrac{1}{\sqrt{n-1}+\sqrt{n-2}}\)

\(=\dfrac{\sqrt{n-1}+\sqrt{n-2}}{\sqrt{n-1}+\sqrt{n}}< 1\)

=>Đây là dãy số giảm

1:

a: \(u_2=2\cdot1+3=5;u_3=2\cdot5+3=13;u_4=2\cdot13+3=29;\)

\(u_5=2\cdot29+3=61\)

b: \(u_2=u_1+2^2\)

\(u_3=u_2+2^3\)

\(u_4=u_3+2^4\)

\(u_5=u_4+2^5\)

Do đó: \(u_n=u_{n-1}+2^n\)

a: \(2n^2-3>=-3\)

\(\Leftrightarrow u_n=\dfrac{1}{2n^2-3}< =-\dfrac{1}{3}\)

=>Dãy số bị chặn trên ở -1/3

b: \(2n^2-1>=-1\)

=>\(u_n=\dfrac{1}{2n^2-1}< =\dfrac{1}{-1}=-1\)

=>Dãy số bị chặn trên ở -1

a:

\(0< =cos\left(\dfrac{\Omega}{2n}\right)< =1;n\in Z^+\)

Khi n chẵn thì \(\left(-1\right)^n=1\)

=>\(u_n=cos\left(\dfrac{\Omega}{2n}\right)\)

=>\(0< =u_n< =1\)

=>\(\left(u_n\right)\) bị chặn ở khoảng [0;1]

Khi n lẻ thì \(\left(-1\right)^n=-1\)

=>\(u_n=-cos\left(\dfrac{\Omega}{2n}\right)\)

\(0< =cos\left(\dfrac{\Omega}{2n}\right)< =1\)

=>\(0>=-cos\left(\dfrac{\Omega}{2n}\right)>=-1\)

=>\(0>=u_n>=-1\)

=>\(\left(u_n\right)\) bị chặn ở khoảng [-1;0]

b: \(-1< =\dfrac{1}{5^n}< =0\)

=>\(-\sqrt{2}< =\dfrac{\sqrt{2}}{5^n}< =0\)

=>\(-\sqrt{2}< =t_n< =0\)

Vậy: Dãy số bị chặn ở khoảng \(\left[-\sqrt{2};0\right]\)

\(u_n=\dfrac{3^n-1}{2^n}\)

\(\Rightarrow u_{n+1}=\dfrac{3^{n+1}-1}{2^{n+1}}\)

\(\Rightarrow u_{n+1}-u_n=\dfrac{3^{n+1}-1}{2^{n+1}}-\dfrac{3^n-1}{2^n}\)

\(\Rightarrow u_{n+1}-u_n=\dfrac{2^n.3^{n+1}-2^n-2^{n+1}.3^n+2^{n+1}}{2^n.2^{n+1}}\)

\(=\dfrac{2^n.3^n\left(3-2\right)-2^n\left(2-1\right)}{2^{2n+1}}\)

\(=\dfrac{2^n.\left(3^n-1\right)}{2^{2n+1}}\)

\(=\dfrac{\left(3^n-1\right)}{2}>0\left(n>1\right)\)

Vậy dãy \(u_n\)đã cho tăng

Bạn xem lại xem viết đề có thiếu/nhầm gì không?

\(u_2=\sqrt{2}\left(2+3\right)-3=5\sqrt{2}-3\)

\(u_3=\sqrt{\dfrac{3}{2}}.5\sqrt{2}-3=5\sqrt{3}-3\)

\(u_4=\sqrt{\dfrac{4}{3}}.5\sqrt{3}-3=5\sqrt{4}-3\)

....

\(\Rightarrow u_n=5\sqrt{n}-3\)

\(\Rightarrow\lim\limits\dfrac{u_n}{\sqrt{n}}=\lim\limits\dfrac{5\sqrt{n}-3}{\sqrt{n}}=5\)

\(u_n=\sqrt[]{n+10}-\sqrt[]{n+2}\)

\(\Leftrightarrow u_n=\dfrac{n+10-\left(n+2\right)}{\sqrt[]{n+10}+\sqrt[]{n+2}}\)

\(\Leftrightarrow u_n=\dfrac{8}{\sqrt[]{n+10}+\sqrt[]{n+2}}\)

\(u_{n+1}=\sqrt[]{n+11}-\sqrt[]{n+3}\)

\(\Leftrightarrow u_{n+1}=\dfrac{n+11-\left(n+3\right)}{\sqrt[]{n+11}+\sqrt[]{n+3}}\)

\(\Leftrightarrow u_{n+1}=\dfrac{8}{\sqrt[]{n+11}+\sqrt[]{n+3}}\)

\(u_{n+1}-u_n=8\left(\dfrac{1}{\sqrt[]{n+11}+\sqrt[]{n+3}}-\dfrac{1}{\sqrt[]{n+10}+\sqrt[]{n+2}}\right)\)

mà \(\dfrac{1}{\sqrt[]{n+11}+\sqrt[]{n+3}}< \dfrac{1}{\sqrt[]{n+10}+\sqrt[]{n+2}}\)

\(\Rightarrow u_{n+1}-u_n< 0\)

Vậy dãy đã cho là dãy số giảm