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:

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

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

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

1: u2=4 và u4=10

=>u1+d=4 và u1+3d=10

=>2d=6 và u1+d=4

=>d=3 và u1=1

\(S_{10}=\dfrac{10\cdot\left(2\cdot1+9\cdot3\right)}{2}=5\cdot\left(2+27\right)=145\)

2: 

u3=6 và u5=16

=>u1+2d=6 và u1+4d=16

=>2d=10  và u1+2d=6

=>d=5 và u1=6-2*5=-4

\(S_{12}=\dfrac{12\cdot\left(2\cdot\left(-4\right)+11\cdot5\right)}{2}=6\cdot\left(-8+55\right)=6\cdot47=282\)

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.

\(\Leftrightarrow\left\{{}\begin{matrix}u_1-u_1-2q+u_1+4q=65\\u_1+u_1+6q=325\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1+2q=65\\2u1+6q=325\end{matrix}\right.\)

=>u1=-130; q=195/2

`u_n = u_1 + (n-1).d`

`{(u_1-u_3+u_5=65),(u_1+u_7=325):}`

`<=>{(u_1-u_1-2d+u_1+4d=65),(u_1+u_1+6d=325):}`

`<=>{(u_1+2d=65),(2u_1+6d=325):}`

`<=>{(u_1=-130),(u_2=195/2):}` 

`u_n=u_1 . q^(n-1)`

`{(u_1-u_3+u_5=65),(u_1+u_7=325):}`

`<=>{(u_1 -u_1 .q^2 +u_1 .q^4=65),(u_1+u_1 .q^6=325):}`

`<=>{(u_1(1- q^2 + q^4)=65 \(1)),(u_1 .(1+q^6=325 \(2)):}`

Lấy (2) : (1) được: `(q^6+1)/(q^4-q^2+1)=5`

`<=>q=+-2`

TH1: `q=2=>u_1=5`

TH2: `q=-2=> u_1=5`

Vậy `(u_1;q)=(5;2),(5;-2)` 

1: u3=-3 và u9=29

=>u1+2d=-3 và u1+8d=29

=>-6d=-32 và u1+2d=-3

=>d=16/3 và u1=-3-2d=-3-32/3=-41/3

2: \(S_{20}=\dfrac{20\cdot\left[2\cdot u1+19\cdot d\right]}{2}=10\cdot\left(-5\cdot2+19\cdot3\right)\)

=10(57-10)

=10*47=470

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/ \(u_6=u_1+5d=8\Rightarrow u_1=8-5d\)

\(u_2=u_1+d;u_4=u_1+3d\)

\(\Rightarrow\left\{{}\begin{matrix}u_2=8-5d+d=8-4d\\u_4=8-5d+3d=8-2d\end{matrix}\right.\)

\(\Rightarrow\left(8-4d\right)^2+\left(8-2d\right)^2=16\Rightarrow...\)

b/ Câu này làm theo ý hiểu thôi, ko chắc đâu

\(Xet-S_n:\)

\(u_1=u_1\)

\(u_2=u_1+d\)

\(u_3=u_1+2d\)

......

\(u_n=u_1+\left(n-1\right)d\)

\(\Rightarrow S_n=u_1+u_2+...+u_n=u_1+u_1+d+...+u_1+\left(n-1\right)d=n.u_1+d+2d+....+\left(n-1\right)d\)

\(=n.u_1+\left(1+2+...+\left(n-1\right)\right)d=n.u_1+\dfrac{d\left(n-1\right).n}{2}=\dfrac{n\left[2u_1+\left(n-1\right)d\right]}{2}\)

Tương tụ với S(2n)

\(S_{2n}=u_1+u_2+...+u_{2n}=u_1+u_1+d+....+u_1+\left(2n-1\right)d\)

\(=2n.u_1+d+2d+...+\left(2n-1\right)d=2n.u_1+\left(1+2+...+\left(2n-1\right)\right)d=2n.u_1+d.n\left(2n-2\right)=2n\left(u_1+\left(n-1\right).d\right)\)

\(4S_n=S_{2n}\Leftrightarrow4.\dfrac{n\left[2u_1+\left(n-1\right)d\right]}{2}=2n\left(u_1+\left(n-1\right).d\right)\)

\(\Leftrightarrow2n\left[2u_1+\left(n-1\right)d\right]=2n\left[u_1+\left(n-1\right)d\right]\)\(\Leftrightarrow2u_1=u_1\Rightarrow u_1=0\)

\(u_5=u_1+4d=18\Rightarrow d=\dfrac{18}{4}=4,5\)

Ok check lại số má hộ tui nhó