a)
\(u_1=10^{1-2.1}=10^{-1};u_2=10^{1-2.2}=10^{-3}\);
\(u_3=10^{1-2.3}=10^{-5}\); \(u_4=10^{1-2.4}=10^{-7}\);
\(u_5=10^{1-2.5}=10^{-9}\).
Xét \(\dfrac{u_n}{u_{n-1}}=\dfrac{10^{1-2n}}{10^{1-2\left(n-1\right)}}=\dfrac{10^{1-2n}}{10^{3-2n}}=10^{-2}=\dfrac{1}{100}\).
Suy ra: \(u_n=\dfrac{1}{100}u_{n-1}\) và dễ thấy \(\left(u_n\right)>0,\forall n\in N^{\circledast}\) nên \(u_n< u_{n-1},\forall n\ge2\).
Vậy \(\left(u_n\right)\) là dãy số tăng.
b) \(u_1=3^1-7=-4\); \(u_2=3^2-7=2\); \(u_3=3^3-7=25\);
\(u_4=3^4-7=74\); \(u_5=3^5-7=236\).
\(u_n-u_{n-1}=3^n-7-\left(3^{n-1}-7\right)=3^n-3^{n-1}=2.3^{n-1}\)\(\left(n\ge2\right)\).
Với \(n\ge2\) thì \(2.3^{n-1}>0\) nên \(u_n>u_{n-1}\).
Vậy \(\left(u_n\right)\) là dãy số tăng.
c)
\(u_1=\dfrac{2.1+1}{1^2}=2;u_2=\dfrac{2.2+1}{2^2}=\dfrac{5}{2}\);
\(u_3=\dfrac{2.3+1}{3^2}=\dfrac{7}{9}\); \(u_4=\dfrac{2.4+1}{4^2}=\dfrac{9}{16}\); \(u_5=\dfrac{2.5+1}{5^2}=\dfrac{11}{25}\);
\(u_n=\dfrac{2n+1}{n^2}=\dfrac{2}{n}+\dfrac{1}{n^2}\);
\(u_{n+1}=\dfrac{2}{n+1}+\dfrac{1}{\left(n+1\right)^2}\);
Vì\(n+1>n\) nên suy ra:
\(\dfrac{2}{n+1}< \dfrac{2}{n}\)
\(\dfrac{1}{\left(n+1\right)^2}< \dfrac{1}{n^2}\)
Cộng từng vế của bất đẳng thức ta có:
\(\dfrac{2}{n+1}+\dfrac{1}{\left(n+1\right)^2}< \dfrac{2}{n}+\dfrac{1}{n^2}\)
\(\Leftrightarrow u_{n+1}< u_n\).
Vậy \(\left(u_n\right)\) là dãy số giảm
d)
\(u_1=\dfrac{3^1.\sqrt{1}}{2^1}=\dfrac{3}{2}\); \(u_2=\dfrac{3^2.\sqrt{2}}{2^2}=\dfrac{9\sqrt{2}}{4}\); \(u_3=\dfrac{3^3.\sqrt{3}}{2^3}=\dfrac{27\sqrt{2}}{8}\); \(u_4=\dfrac{3^4.\sqrt{4}}{2^4}=\dfrac{81}{8}\); \(u_5=\dfrac{3^5.\sqrt{5}}{2^5}=\dfrac{243\sqrt{2}}{32}\).
Xét:
\(\dfrac{u_n}{u_{n-1}}=\dfrac{3^n\sqrt{n}}{2^n}:\dfrac{3^{n-1}\sqrt{n-1}}{2^{n-1}}\)\(=\left(\dfrac{3}{2}\right)^{n-\left(n-1\right)}.\dfrac{\sqrt{n}}{\sqrt{n-1}}\)
\(=\dfrac{3}{2}\sqrt{\dfrac{n}{n-1}}>\dfrac{3}{2}\sqrt{\dfrac{n}{n+n}}=\dfrac{3}{2}\sqrt{\dfrac{1}{2}}=\dfrac{3\sqrt{2}}{4}>1\).
Dễ thấy \(\left(u_n\right)\) là dãy số không âm nên \(u_n>u_{n-1}\).
Vậy \(\left(u_n\right)\) là dãy số tăng.
