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:

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

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

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

=-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

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:

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

a: u1-2u4+u6=12 và u2+u5=8

=>u1-2u1-6d+u1+5d=12 và u1+d+u1+4d=8

=>d=12 và 2u1+5d=8

=>d=12 và 2u1=8-5d=8-60=-52

=>u1=-26 và d=12

b: u5-u2=3 và u3*u8=24

=>u1+4d-u1-d=3 và (u1+2d)(u1+7d)=24

=>d=1 và (u1+2)(u1+7)=24

=>d=1 và u1^2+9u1-10=0

=>d=1 và (u1=-10 hoặc u1=1)