Câu 1:

\(S_8=u_1+u_2+u_3+...+u_8\)

\(=\dfrac{u_1\left(1-q^8\right)}{1-q}=\dfrac{2048\cdot\left(1-\left(\dfrac{5}{4}\right)^8\right)}{1-\dfrac{5}{4}}\)

\(=\dfrac{325089}{8}\)

2: \(S_{10}=u_1+u_2+...+u_9+u_{10}\)

=>\(S_{10}=\dfrac{u_1\left(1-q^{10}\right)}{1-q}=\dfrac{-3\cdot\left(1-\left(\dfrac{1}{2}\right)^{10}\right)}{1-\dfrac{1}{2}}\)

\(=-6\cdot\left(1-\dfrac{1}{2^{10}}\right)=-6+\dfrac{6}{2^{10}}=-\dfrac{3069}{512}\)

1:

\(S_8=\dfrac{u_1\cdot\left(1-q^8\right)}{1-q}=\dfrac{2048\cdot\left(1-\left(\dfrac{5}{4}\right)^8\right)}{1-\dfrac{5}{4}}\)

\(=-8192\left(1-\left(\dfrac{5}{4}\right)^8\right)\)

2:

\(u2=u1\cdot q\)

=>\(q=\dfrac{3}{-1}=-3\)

\(S_{10}=\dfrac{u1\left(1-q^{10}\right)}{1-q}=\dfrac{-1\cdot\left(1-\left(-3\right)^{10}\right)}{1-\left(-3\right)}\)

\(=\dfrac{-1}{4}\left(1-3^{10}\right)\)

\(Bài.1:u_n=\dfrac{3}{2}.\left(\dfrac{1}{2}\right)^n=\dfrac{3}{512}\\ \Rightarrow\left(\dfrac{1}{2}\right)^n=\dfrac{3}{512}:\dfrac{3}{2}=\dfrac{1}{256}=\dfrac{1}{2^8}\\ Mà:\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{2}\right)^8\\ Vậy:n=8\\ \Rightarrow Vậy:\dfrac{3}{512}.là.số.hạng.thứ.8\)

\(\dfrac{2}{y-x};\dfrac{1}{y};\dfrac{2}{y-z}\) là cấp số cộng

\(\Rightarrow\dfrac{2}{y-x}+\dfrac{2}{y-z}=\dfrac{2}{y}\)

\(\Leftrightarrow2y^2-2xz=0\Leftrightarrow y^2=xz\)

=> x,y,x là cấp số nhân

I.

Do \(\left(u_n\right)\) là cấp số nhân \(\Rightarrow\)\(u_4=u_3.q\Rightarrow q=\dfrac{u_4}{u_3}=\dfrac{10}{3}\)

\(u_3=u_1q^2\Rightarrow u_1=\dfrac{u_3}{q^2}=\dfrac{27}{100}\)

2. Công thức số hạng tổng quát: \(u_n=\dfrac{27}{100}.\left(\dfrac{10}{3}\right)^{n-1}\)

II.

1. \(\lim\limits\dfrac{-3n^2+2n-2022}{3n^2-2022}=\lim\dfrac{-3+\dfrac{2}{n}-\dfrac{2022}{n^2}}{3-\dfrac{2022}{n^2}}=\dfrac{-3+0-0}{3-0}=-1\)

2.

\(\lim\limits_{x\rightarrow2}\dfrac{x^2-5x+6}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x-3\right)}{x-2}=\lim\limits_{x\rightarrow2}\left(x-3\right)=-1\)

1/ \(=\lim\limits_{x\rightarrow-\infty}x\left(-\sqrt{\dfrac{16x^2}{x^2}-\dfrac{3x}{x^2}+\dfrac{5}{x^2}}+2-\dfrac{5}{x}\right)=\lim\limits_{x\rightarrow-\infty}x\left(-4+2\right)=-\infty\)

\(=\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{\dfrac{16x^2}{x^2}-\dfrac{3x}{x^2}+\dfrac{5}{x^2}}+2-\dfrac{5}{x}\right)=\lim\limits_{x\rightarrow+\infty}x\left(4+2\right)=+\infty\)

2/ \(S=\dfrac{-\dfrac{1}{3}}{1+\dfrac{1}{3}}=-\dfrac{1}{4}\)

4/ 

5/ 

\(f'\left(x\right)=4\left(2m-1\right)x^3-4x\)

Vì tiếp tuyến vuông góc với \(y=5x-2018\Rightarrow f'\left(x\right)=-\dfrac{1}{5}\)

\(\Rightarrow f'\left(1\right)=-\dfrac{1}{5}\Leftrightarrow4\left(2m-1\right)-4=-\dfrac{1}{5}\Leftrightarrow m=\dfrac{39}{40}\)

a) Với mọi ∀n ε N^*, ta có ( . 2ⁿ⁺¹) : ( . 2ⁿ) = 2.

Suy ra u_n+1 = u_n.2, với n ε N^*

Vậy dãy số đã chp là một câp số nhân với u₁ = , q = 2.

b) Với mọi ∀n ε N^*, ta có u_n+1 = =u_n.

Vậy dãy số đã cho là một cấp số nhân với u₁ = , q =

c) Với mọi ∀n ε N^*, ta có u_n+1 = .

Cấp số nhân lùi vô hạn có \(u_1=1\) và \(q=\dfrac{1}{2}\)

\(\Rightarrow S=\dfrac{u_1}{1-q}=\dfrac{1}{1-\dfrac{1}{2}}=2\)

1:

\(S_{10}=\dfrac{u_1\cdot\left(1-q^{10}\right)}{1-q}=\dfrac{-3\cdot\left(1-\dfrac{1}{1024}\right)}{1-\dfrac{1}{2}}\)

\(=-6\cdot\dfrac{1023}{1024}=\dfrac{-3069}{512}\)

2:

\(\left\{{}\begin{matrix}u1=6\\u2=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u1=6\\u1\cdot q=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u1=6\\q=3\end{matrix}\right.\)

\(S_{12}=\dfrac{u_1\left(1-q^{12}\right)}{1-q}=\dfrac{6\cdot\left(1-3^{12}\right)}{1-3}=-3\cdot\left(1-3^{12}\right)\)

\(=3^{13}-3\)

\(\left\{{}\begin{matrix}u_1=\dfrac{1}{2}\\q=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow S=\dfrac{1}{2}.\dfrac{1}{1+\dfrac{1}{2}}=\dfrac{1}{3}\)