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:

\(u2=u1\cdot q\)

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

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

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

1: \(S_{99}=\dfrac{99\cdot\left[2\cdot6+98\cdot\left(-2\right)\right]}{2}=99\cdot\left(6-98\right)\)

=-9108

2: \(S_{100}=\dfrac{100\cdot\left(2\cdot\left(-2\right)+99\cdot4\right)}{2}=50\left(-4+99\cdot4\right)\)

=50*392

=19600

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 

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

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:

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

Ta có:

\(u_2=u_1.q\\ u_3=u_2.q=\left(u_1.q\right).q=u_1.q^2\\ u_4=u_3.q=\left(u_1.q^2\right).q=u_1.q^3.\\ .\\ .\\ .\\ u_{10}=u_1.q^9\)

\(a,u_1+u_n=u_1+\left[u_1+\left(n-1\right)d\right]=u_1+u_1+\left(n-1\right)d=2u_1+\left(n-1\right)d\\ u_2+u_{n-1}=\left[u_1+d\right]+\left[u_1+\left(n-2\right)d\right]=2u_1+\left(n-1\right)d\\ ...\\ u_k+u_{n-k+1}=\left[u_1+\left(k-1\right)d\right]+\left[u_1+\left(n-k+1-1\right)d\right]=2u_1+\left(n-1\right)d\)

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

Cộng vế với vế, ta được:

\(2\left(u_1+u_2+...+u_n\right)=n\left[2u_1+\left(n-1\right)d\right]\\ \Leftrightarrow2\left(u_1+u_2+...+u_n\right)=n\left(u_1+u_n\right)\)