Câu hỏi của Ngoc Chau

Question

1) Tìm số hạng đầu u1 và công bội q của Cấp Số Nhân (Un)

a) Biết u2=-10, u3=-20

b) Biết u3=2, u6=1/4

c) Biết{ u1-u3=-9; u3-u5=-36}

2)Tìm S8 của Cấp Số Nhân (Un) biết:

{u4-u2=24; u3-u1=12}

3) Cho...

Nguyễn Việt Lâm · Accepted Answer

1.

$\left\{{}\begin{matrix}u_2=u_1q=-10\u_3=u_1q^2=-20\end{matrix}\right.$ $\Rightarrow\frac{u_1q^2}{u_1q}=\frac{-20}{-10}\Rightarrow q=2$

$\Rightarrow u_1=\frac{-10}{q}=-5$

b/ $\left\{{}\begin{matrix}u_1q^2=2\u_1q^5=\frac{1}{4}\end{matrix}\right.$ $\Rightarrow q^3=\frac{1}{8}\Rightarrow q=\frac{1}{2}$

$\Rightarrow u_1=\frac{2}{q^2}=8$

c/ $\left\{{}\begin{matrix}u_1-u_1q^2=-9\u_1q^2-u_1q^4=-36\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}u_1\left(1-q^2\right)=-9\u_1q^2\left(1-q^2\right)=-36\end{matrix}\right.$

$\Leftrightarrow\frac{u_1q^2\left(1-q^2\right)}{u_1\left(1-q^2\right)}=\frac{-36}{-9}\Rightarrow q^2=4\Rightarrow\left[{}\begin{matrix}q=2\q=-2\end{matrix}\right.$

$\Rightarrow u_1=\frac{-9}{1-q^2}=\frac{-9}{-3}=3$Câu trả lời: 1.

$\left\{{}\begin{matrix}u_2=u_1q=-10\u_3=u_1q^2=-20\end{matrix}\right.$ $\Rightarrow\frac{u_1q^2}{u_1q}=\frac{-20}{-10}\Rightarrow q=2$

$\Rightarrow u_1=\frac{-10}{q}=-5$

b/ $\left\{{}\begin{matrix}u_1q^2=2\u_1q^5=\frac{1}{4}\end{matrix}\right.$ $\Rightarrow q^3=\frac{1}{8}\Rightarrow q=\frac{1}{2}$

$\Rightarrow u_1=\frac{2}{q^2}=8$

c/ $\left\{{}\begin{matrix}u_1-u_1q^2=-9\u_1q^2-u_1q^4=-36\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}u_1\left(1-q^2\right)=-9\u_1q^2\left(1-q^2\right)=-36\end{matrix}\right.$

$\Leftrightarrow\frac{u_1q^2\left(1-q^2\right)}{u_1\left(1-q^2\right)}=\frac{-36}{-9}\Rightarrow q^2=4\Rightarrow\left[{}\begin{matrix}q=2\q=-2\end{matrix}\right.$

$\Rightarrow u_1=\frac{-9}{1-q^2}=\frac{-9}{-3}=3$

Nguyễn Việt Lâm · Answer

Bài 2:

$\left\{{}\begin{matrix}u_1q^3-u_1q=24\u_1q^2-u_1=12\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}u_1q\left(q^2-1\right)=24\u_1\left(q^2-1\right)=12\end{matrix}\right.$

$\Leftrightarrow\frac{u_1q\left(q^2-1\right)}{u_1\left(q^2-1\right)}=\frac{24}{12}\Rightarrow q=2\Rightarrow u_1=\frac{12}{q^2-1}=4$

$\Rightarrow S_8=u_1.\frac{q^8-1}{q-1}=4\left(2^8-1\right)=...$

Câu 3:

$u_{10}=u_1q^9=4\left(-2\right)^9=-2^{11}$

$S_{15}=u_1.\frac{q^{15}-1}{q-1}=4.\frac{\left(-2\right)^{15}-1}{-3}=\frac{3}{4}\left(2^{15}+1\right)$Câu trả lời: Bài 2:

$\left\{{}\begin{matrix}u_1q^3-u_1q=24\u_1q^2-u_1=12\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}u_1q\left(q^2-1\right)=24\u_1\left(q^2-1\right)=12\end{matrix}\right.$

$\Leftrightarrow\frac{u_1q\left(q^2-1\right)}{u_1\left(q^2-1\right)}=\frac{24}{12}\Rightarrow q=2\Rightarrow u_1=\frac{12}{q^2-1}=4$

$\Rightarrow S_8=u_1.\frac{q^8-1}{q-1}=4\left(2^8-1\right)=...$

Câu 3:

$u_{10}=u_1q^9=4\left(-2\right)^9=-2^{11}$

$S_{15}=u_1.\frac{q^{15}-1}{q-1}=4.\frac{\left(-2\right)^{15}-1}{-3}=\frac{3}{4}\left(2^{15}+1\right)$

Nguyễn Việt Lâm · Answer

Bài 4:

$u_n=5.\left(\frac{1}{2}\right)^{2n-1}=10.\left(\frac{1}{2}\right)^{2n}=10\left(\frac{1}{4}\right)^n$

Là cấp số nhân với $u_1=10$ và công bội $q=\frac{1}{4}$

Bài 5:

$S_5=u_1.\frac{q^4-1}{q-1}=u_1.\frac{\left(\frac{1}{3}\right)^4-1}{\frac{1}{3}-1}=\frac{121}{81}u_1$

$\Rightarrow u_1=\frac{81}{121}S_5=81$

Bài 6:

$\left\{{}\begin{matrix}u_1q=4\u_1q^3=9\end{matrix}\right.$ $\Rightarrow\left(u_1q^2\right)^2=36\Rightarrow\left[{}\begin{matrix}u_1q^2=6\u_1q^2=-6\end{matrix}\right.$

Mà $u_3=u_1q^2\Rightarrow u_3=\pm6$Câu trả lời: Bài 4: