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:

\(u2=u1\cdot q\)

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

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

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

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, Ta có \(\left\{{}\begin{matrix}u_1=-1\\u_1.q=3\end{matrix}\right.\Rightarrow\dfrac{1}{q}=-\dfrac{1}{3}\Leftrightarrow q=-3\)

\(S_{10}=-1.\dfrac{1-\left(-3\right)^{10}}{1-\left(-3\right)}=14762\)

2, tương tự 

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:

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

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

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

1) \(\left(u_n\right):\left\{{}\begin{matrix}u_1=-7\\q=2\end{matrix}\right.\)

\(u_5=-7.q^4=-7.16=-112\)

\(u_m=u_1.q^{m-1}\)

\(\Leftrightarrow-7.2^{m-1}=-3584\)

\(\Leftrightarrow2^{m-1}=512=2^9\)

\(\Leftrightarrow m-1=9\)

\(\Leftrightarrow m=10\)

Vậy số \(-3584\) là số thứ \(10\) của cấp số nhân

\(\left(u_n\right):\left\{{}\begin{matrix}u_1=-3\\q=-2\end{matrix}\right.\)

\(u_{10}=-u_1.q^9=-3.\left(-2\right)^9=1536\)

\(u_m=u_1.q^{m-1}\)

\(\Leftrightarrow-3.\left(-2\right)^{m-1}=-3072\)

\(\Leftrightarrow\left(-2\right)^{m-1}=1024=\left(-2\right)^{10}\)

\(\Leftrightarrow m-1=10\)

\(\Leftrightarrow m=11\)

Vậy số \(-3072\) là số thứ \(11\) của cấp số nhân.

\(Bài.1:\\ u_7=u_1+6d\\ \Leftrightarrow-10=2+6d\\ \Rightarrow6d=-10-2=-12\\ Vậy:d=\dfrac{-12}{6}=-2\\ Bài.2:S_{10}=10.u_1+\dfrac{10.\left(10-1\right)}{2}.d=10.1+\dfrac{10.9}{2}.2=100\\ Bài.3:S_{2019}=2019.u_1+\dfrac{2019.\left(2019-1\right)}{2}.d\\ =2019.3+\dfrac{2019.2018}{2}.2=2019.2021=4080399\)

Bài 4:

\(d=u_2=u_1=5-2=3\)

Bài 5:

\(u_n=u_1+\left(n-1\right)d\\ \Leftrightarrow2018=2+\left(n-1\right).9\\ \Leftrightarrow2+9n-9=2018\\ \Leftrightarrow9n=2018-2+9\\ \Leftrightarrow9n=2025\\ \Leftrightarrow n=\dfrac{2025}{9}=225\)

Vậy: 2018 là số hạng thứ 225 của dãy

Bài 6:

Đề chưa có yêu cầu

4: d=u2-u1=3

5: Đặt 2018=2+(n-1)*9

=>9(n-1)=2016

=>n-1=224

=>n=225

=>2018 là số thứ 225

3:

\(S_{2019}=2019\left(\dfrac{2\cdot3+2018\cdot2}{2}\right)=4080399\)

2:

\(S_{10}=\dfrac{10\cdot\left(2\cdot1+9\cdot2\right)}{2}=10\left(1+9\right)=100\)

a) \(\left| q \right| = \left| {\frac{1}{2}} \right| < 1\)

b) \(\begin{array}{l}{S_n} = {u_1} + {u_2} + ... + {u_n} = {u_1}.\frac{{1 - {q^n}}}{{1 - q}} = 1.\frac{{1 - {{\left( {\frac{1}{2}} \right)}^n}}}{{1 - \frac{1}{2}}} = 2 - 2.{\left( {\frac{1}{2}} \right)^n}\\ \Rightarrow \lim {S_n} = \lim \left[ {2 - 2.{{\left( {\frac{1}{2}} \right)}^n}} \right] = \lim 2 - 2\lim {\left( {\frac{1}{2}} \right)^n} = 2\end{array}\)

a) Ta có:

\(q.{S_n} = q.\left( {{u_1} + {u_2} + ... + {u_n}} \right) = {u_1}.q + {u_2}.q + ... + {u_n}.q = \left( {{u_2} + {u_3} + ... + {u_n}} \right) + q.{u_n}\)

b) Ta có:

\({u_1} + q.{S_n} = {u_1} + \left( {{u_2} + {u_3} + ... + {u_n}} \right) + q.{u_n} = \left( {{u_1} + {u_2} + {u_3} + ... + {u_n}} \right) + q.{u_n} = {S_n} + {u_1}.{q^n}\)

1. Gọi công bội của csn đó là $q$ thì:
$u_6=q^4u_2$

$\Leftrightarrow 32=q^4.2\Leftrightarrow q^4=16$

$\Leftrightarrow q=\pm 2$

2. 

$u_{2019}=q^{2018}u_1=2.3^{2018}$

a)    Ta có:

-        Số hạng thứ nhất: \({u_1}\)

-        Số hạng thứ hai: \({u_2} = {u_1}.q\)

-        Số hạng thứ ba: \({u_3} = {u_2}.q = \left( {{u_1}.q} \right).q = {u_1}.{q^2}\)

-        Số hạng thứ tư: \({u_4} = {u_3}.q = \left( {{u_1}.{q^2}} \right).q = {u_1}.{q^3}\)

-        Số hạng thứ năm: \({u_5} = {u_4}.q = \left( {{u_1}.{q^3}} \right).q = {u_1}.{q^4}\)

b)    Dự đoán công thức tính: \({u_n} = {u_1}.{q^{n - 1}}\)