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Chọn A.
Dễ thấy un là cấp số nhân với q = 10
Ta có: u8 = 107u1; u10 = 109u1
Do đó PT 
Giải PT ta được logu1 = -17 ⇔ u1 = 10-17 ⇒ u2018 = 102017 u1 = 102000
Câu 1.
Vì \(\sqrt{2},\left(\sqrt{2}\right)^2,...,\left(\sqrt{2}\right)^n\) lập thành cấp số nhân có \(u_1=\sqrt{2}=q\) nên
\({u_n} = \sqrt 2 .\dfrac{{1 - {{\left( {\sqrt 2 } \right)}^n}}}{{1 - \sqrt 2 }} = \left( {2 - \sqrt 2 } \right)\left[ {{{\left( {\sqrt 2 } \right)}^n} - 1} \right] \to \lim {u_n} = + \infty \) vì \(\left\{{}\begin{matrix}a=2-\sqrt{2}>0\\q=\sqrt{2}>1\end{matrix}\right.\)
Câu 3.
Ta có biến đổi:
\(\lim \left( {\dfrac{{{n^2} - n}}{{1 - 2{n^2}}} + \dfrac{{2\sin {n^2}}}{{\sqrt n }}} \right) = \lim \dfrac{{{n^2} - n}}{{1 - 2{n^2}}} = \dfrac{1}{2}\)
Câu 4.
\(\lim \left( {{n^2}\sin \dfrac{{n\pi }}{5} - 2{n^3}} \right) = \lim {n^3}\left( {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n} - 2} \right) = - \infty \)
Vì \(\lim {n^3} = + \infty ;\lim \left( {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n} - 2} \right) = - 2 \)
\(\left| {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n}} \right| \le \dfrac{1}{n};\lim \dfrac{1}{n} = 0 \Rightarrow \lim \left( {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n} - 2} \right) = - 2\)
Câu 5.
Ta có: \(\left\{ \begin{array}{l} 0 \le \left| {{u_n}} \right| \le \dfrac{1}{{{n^2} + 1}} \le \dfrac{1}{n} \to 0\\ 0 \le \left| {{v_n}} \right| \le \dfrac{1}{{{n^2} + 2}} \le \dfrac{1}{n} \to 0 \end{array} \right. \to \lim {u_n} = \lim {v_n} = 0 \to \lim \left( {{u_n} + {v_n}} \right) = 0\)
a/
\(u_n=\dfrac{1}{\left(2-1\right)\left(2+1\right)}+\dfrac{1}{\left(3-1\right)\left(3+1\right)}+...+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)
\(u_n=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(n-2\right)n}+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)
\(u_n=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n-2}-\dfrac{1}{n}+\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)
\(u_n=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{1}{2}\left(\dfrac{3}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
\(\Rightarrow lim\left(u_n\right)=lim\left(\dfrac{1}{2}\left(\dfrac{3}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\right)=\dfrac{1}{2}.\dfrac{3}{2}=\dfrac{3}{4}\)
b/ \(u_n=\dfrac{1}{1^2+3}+\dfrac{1}{2^2+6}+...+\dfrac{1}{n^2+3n}=\dfrac{1}{1.4}+\dfrac{1}{2.5}+...+\dfrac{1}{n\left(n+3\right)}\)
\(u_n=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{n}-\dfrac{1}{n+3}\right)\)
\(u_n=\dfrac{1}{3}\left(1+\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{n+1}-\dfrac{1}{n+2}-\dfrac{1}{n+3}\right)\)
\(\Rightarrow lim\left(u_n\right)=lim\left(\dfrac{1}{3}\left(1+\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{n+1}-\dfrac{1}{n+2}-\dfrac{1}{n+3}\right)\right)\)
\(\Rightarrow lim\left(u_n\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)=\dfrac{11}{18}\)
\(u_3+u_7+...+u_{35}=u_1q^2+u_1q^6+...+u_1q^{34}\)
\(=u_1q^2\left(1+q^4+q^8+...+q^{32}\right)=u_1q^2.\frac{\left(q^4\right)^9-1}{q^4-1}=524286\)
2/ \(u_1^2+u_2^2+...+u_{20}^2=u_1^2+u_1^2q^2+u_1^2q^4+...+u_1^2q^{38}\)
\(=u_1^2\left(1+q^2+q^4+...+q^{38}\right)=u_1^2\frac{\left(q^2\right)^{20}-1}{q^2-1}=\frac{3^{20}-1}{2}\)
3/
\(u_1=2;u_n=18\)
\(u_1^2+u_2^2+...+u_n^2=484\)
\(\Leftrightarrow u_1^2+u_1^2q^2+...+u_1^2q^{2\left(n-1\right)}=484\)
\(\Leftrightarrow u_1^2\left(1+q^2+...+q^{2\left(n-1\right)}\right)=484\)
\(\Leftrightarrow1+q^2+...+q^{2\left(n-1\right)}=121\)
\(\Leftrightarrow\frac{q^{2n}-1}{q^2-1}=121\)
Mà \(u_n=u_1q^{n-1}\Rightarrow q^{n-1}=\frac{u_n}{u_1}=9\Rightarrow q^n=9q\Rightarrow q^{2n}=81q^2\)
\(\Rightarrow\frac{81q^2-1}{q^2-1}=121\Rightarrow81q^2-1=121q^2-121\)
\(\Rightarrow q^2=3\Rightarrow q=\pm\sqrt{3}\)
Chọn D
Ta có u n là cấp số nhân công bội bằng 2.
Ta có
2 4 u 1 + 1 + 2 3 - 2 u 2 = 2 4 u 1 + 1 + 2 3 - 4 u 1 ≥ 2 2 4 u 1 + 1 . 2 3 - 4 u 1 = 8
Mặt khác
Suy ra
Vậy 2 4 u 1 + 1 + 2 3 - 2 u 2 = 8 log 2 ( 2 u 3 2 - 8 u 2 + 4 )
suy ra giá trị nhỏ nhất của n cần tìm là 2021