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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: \(u_n=u_{n-1}+2^n\)
\(u_n=\dfrac{3^n-1}{2^n}\)
\(\Rightarrow u_{n+1}=\dfrac{3^{n+1}-1}{2^{n+1}}\)
\(\Rightarrow u_{n+1}-u_n=\dfrac{3^{n+1}-1}{2^{n+1}}-\dfrac{3^n-1}{2^n}\)
\(\Rightarrow u_{n+1}-u_n=\dfrac{2^n.3^{n+1}-2^n-2^{n+1}.3^n+2^{n+1}}{2^n.2^{n+1}}\)
\(=\dfrac{2^n.3^n\left(3-2\right)-2^n\left(2-1\right)}{2^{2n+1}}\)
\(=\dfrac{2^n.\left(3^n-1\right)}{2^{2n+1}}\)
\(=\dfrac{\left(3^n-1\right)}{2}>0\left(n>1\right)\)
Vậy dãy \(u_n\)đã cho tăng
\(u_n=\sqrt[]{n+10}-\sqrt[]{n+2}\)
\(\Leftrightarrow u_n=\dfrac{n+10-\left(n+2\right)}{\sqrt[]{n+10}+\sqrt[]{n+2}}\)
\(\Leftrightarrow u_n=\dfrac{8}{\sqrt[]{n+10}+\sqrt[]{n+2}}\)
\(u_{n+1}=\sqrt[]{n+11}-\sqrt[]{n+3}\)
\(\Leftrightarrow u_{n+1}=\dfrac{n+11-\left(n+3\right)}{\sqrt[]{n+11}+\sqrt[]{n+3}}\)
\(\Leftrightarrow u_{n+1}=\dfrac{8}{\sqrt[]{n+11}+\sqrt[]{n+3}}\)
\(u_{n+1}-u_n=8\left(\dfrac{1}{\sqrt[]{n+11}+\sqrt[]{n+3}}-\dfrac{1}{\sqrt[]{n+10}+\sqrt[]{n+2}}\right)\)
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
\(u_2=\sqrt{2}\left(2+3\right)-3=5\sqrt{2}-3\)
\(u_3=\sqrt{\dfrac{3}{2}}.5\sqrt{2}-3=5\sqrt{3}-3\)
\(u_4=\sqrt{\dfrac{4}{3}}.5\sqrt{3}-3=5\sqrt{4}-3\)
....
\(\Rightarrow u_n=5\sqrt{n}-3\)
\(\Rightarrow\lim\limits\dfrac{u_n}{\sqrt{n}}=\lim\limits\dfrac{5\sqrt{n}-3}{\sqrt{n}}=5\)
\(u_n=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{n\left(n+1\right)}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(=1-\dfrac{1}{n+1}< 1\)
=>Hàm số bị chặn trên tại \(u_n=1\)
\(n+1>=1\)
=>\(\dfrac{1}{n+1}< =1\)
=>\(-\dfrac{1}{n+1}>=-1\)
=>\(1-\dfrac{1}{n+1}>=-1+1=0\)
=>Hàm số bị chặn dưới tại 0
\(u_n=1-\dfrac{1}{n+1}=\dfrac{n+1-1}{n+1}=\dfrac{n}{n+1}\)
\(\dfrac{u_n}{u_{n+1}}=\dfrac{n}{n+1}:\dfrac{n+1}{n+2}=\dfrac{n^2+2n}{n^2+2n+1}< 1\)
=>(un) là dãy số tăng
1/ \(\lim\limits\dfrac{\dfrac{2^n}{7^n}-5.7.\left(\dfrac{7}{7}\right)^n}{\dfrac{2^n}{7^n}+\left(\dfrac{7}{7}\right)^n}=-35\)
2/ \(\lim\limits\dfrac{\dfrac{3^n}{7^n}-2.5.\left(\dfrac{5}{7}\right)^n}{\dfrac{2^n}{7^n}+\dfrac{7^n}{7^n}}=0\)
3/ \(\lim\limits\sqrt[3]{\dfrac{\dfrac{5}{n}-\dfrac{8n}{n}}{\dfrac{n}{n}+\dfrac{3}{n}}}=\sqrt[3]{-8}=-2\)
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