Lời giải:
Xét csn $(u_n)$ với công bội $q$

Ta có:

$S_n=u_1+u_2+...+u_n=u_1+u_1q+u_1q^2+....+u_1q^{n-1}$

$=u_1(1+q+q^2+....+q^{n-1})$

$qS_n=u_1(q+q^2+q^3+....+q^n)$

$\Rightarrow qS_n-S_n=u_1(q^n-1)$

$\Rightarrow S_n(q-1)=u_1(q^n-1)$

$\Rightarrow S_n=\frac{u_1(q^n-1)}{q-1}=\frac{u_1(1-q^n)}{1-q}$

Ta có đpcm.

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:

\(u2=u1\cdot q\)

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

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

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

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:

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

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

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

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: \(S_{10}=u_1+u_2+...+u_9+u_{10}\)

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

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

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

\(S=\dfrac{\dfrac{1}{u_1}\left[1-\left(\dfrac{1}{2}\right)^{2020}\right]}{1-\dfrac{1}{2}}=\dfrac{2\left(2^{2020}-1\right)}{2^{2020}u_1}\\ P=\left(u_1+u_2+...+u_{2020}\right)+\left(u_2+u_3+...+u_{2021}\right)\\ =\left(1+q\right)\left(u_1+u_2+...+u_{2020}\right)=3u_1\left(2^{2020}-1\right)\\ \rightarrow SP=\dfrac{3\left(2^{2020}-1\right)^2}{2^{2019}}\)

a) \({u_2} = {u_1}.q\)

\({u_3} = {u_1}.{q^2}\)

…

\({u_{n - 1}} = {u_1}.{q^{n - 2}}\)

\({u_n} = {u_1}.{q^{n - 1}}\)

\({S_n} = {u_1} + {u_1}q +  \ldots  + {u_1}{q^{n - 2}} + {u_1}{q^{n - 1}}\)

b) \(q{S_n} = q{u_1} + {u_1}{q^2} +  \ldots  + {u_1}{q^{n - 1}} + {u_1}{q^n}\)

c) \({S_n} - q{S_n} = \left( {{u_1} + {u_1}q +  \ldots  + {u_1}{q^{n - 2}} + {u_1}{q^{n - 1}}} \right) - (q{u_1} + {u_1}{q^2} +  \ldots  + {u_1}{q^{n - 1}} + {u_1}{q^n})\).

\(\begin{array}{l} \Leftrightarrow \left( {1 - q} \right){S_n} = {u_1} - {u_1}{q^n} = {u_1}\left( {1 - {q^n}} \right)\\ \Rightarrow {S_n} = \frac{{{u_1}\left( {1 - {q^n}} \right)}}{{1 - q}}\end{array}\)