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a) \({u_2} = {u_1}.q\)
\({u_3} = {u_2}.q = {u_1}.{q^2}\)
\({u_4} = {u_3}.q = {u_1}.{q^3}\)
\({u_5} = {u_4}.q = {u_1}.{q^4}\)
b) Từ a suy ra: \({u_n} = {u_1} \times {q^{n - 1}}\).
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: \(S_{10}=u_1+u_2+...+u_9+u_{10}\)
=>\(S_{10}=\dfrac{u_1\left(1-q^{10}\right)}{1-q}=\dfrac{-3\cdot\left(1-\left(\dfrac{1}{2}\right)^{10}\right)}{1-\dfrac{1}{2}}\)
\(=-6\cdot\left(1-\dfrac{1}{2^{10}}\right)=-6+\dfrac{6}{2^{10}}=-\dfrac{3069}{512}\)
a:
ĐKXĐ: \(q\notin\left\{0;1;-1\right\}\)
\(HPT\Leftrightarrow\left\{{}\begin{matrix}u1\cdot q^4-u1=15\\u1\cdot q^3-u1\cdot q=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{q^4-1}{q^3-q}=\dfrac{15}{6}=\dfrac{5}{2}\\u1\left(q^4-1\right)=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2q^4-2=5q^3-5q\\u1\left(q^4-1\right)=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2q^4-5q^3+5q-2=0\\u1\left(q^4-1\right)=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(q-2\right)\left(q-1\right)\left(q+1\right)\left(2q-1\right)=0\\u1\left(q^4-1\right)=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}q=2\\q=\dfrac{1}{2}\end{matrix}\right.\\u1\left(q^4-1\right)=15\end{matrix}\right.\)
TH1: q=2
=>\(u1=\dfrac{15}{2^4-1}=\dfrac{15}{15}=1\)
TH2: q=1/2
=>\(u1=\dfrac{15}{\dfrac{1}{16}-1}=15:\dfrac{-15}{16}=-16\)
b:
\(HPT\Leftrightarrow\left\{{}\begin{matrix}u1-u1\cdot q^2+u1\cdot q^4=65\\u1+u1\cdot q^6=325\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{q^4-q^2+1}{q^6+1}=\dfrac{1}{5}\\u1\left(1+q^6\right)=325\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{q^2+1}=\dfrac{1}{5}\\u1\left(q^6+1\right)=325\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}q^2=4\\u1\left(q^6+1\right)=325\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}q\in\left\{2;-2\right\}\\u1\left(q^6+1\right)=325\end{matrix}\right.\Leftrightarrow u1=\dfrac{325}{65}=5\)
c: \(HPT\Leftrightarrow\left\{{}\begin{matrix}u1\cdot q^3+u1\cdot q^5=-540\\u1\cdot q+u1\cdot q^3=-60\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{q^5+q^3}{q^3+q}=9\\u1\left(q+q^3\right)=-60\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}q^2=9\\u1\left(q+q^3\right)=-60\end{matrix}\right.\)
TH1: q=3
\(u1=-\dfrac{60}{3+3^3}=-\dfrac{60}{30}=-2\)
TH2: q=-3
=>\(u1=-\dfrac{60}{-3-27}=\dfrac{60}{30}=2\)
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:
\(\left\{{}\begin{matrix}u1=6\\u2=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u1=6\\u1\cdot q=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u1=6\\q=3\end{matrix}\right.\)
\(S_{12}=\dfrac{u_1\left(1-q^{12}\right)}{1-q}=\dfrac{6\cdot\left(1-3^{12}\right)}{1-3}=-3\cdot\left(1-3^{12}\right)\)
\(=3^{13}-3\)
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: \(S_{99}=\dfrac{99\cdot\left[2\cdot6+98\cdot\left(-2\right)\right]}{2}=99\cdot\left(6-98\right)\)
=-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
\(a,u_1+u_n=u_1+\left[u_1+\left(n-1\right)d\right]=u_1+u_1+\left(n-1\right)d=2u_1+\left(n-1\right)d\\ u_2+u_{n-1}=\left[u_1+d\right]+\left[u_1+\left(n-2\right)d\right]=2u_1+\left(n-1\right)d\\ ...\\ u_k+u_{n-k+1}=\left[u_1+\left(k-1\right)d\right]+\left[u_1+\left(n-k+1-1\right)d\right]=2u_1+\left(n-1\right)d\)
\(b,u_1+u_n=2u_1+\left(n-1\right)d\\ u_2+u_{n-1}=2u_1+\left(n-1\right)d\\ ...\\ u_n+u_1=2u_1+\left(n-1\right)d\)
Cộng vế với vế, ta được:
\(2\left(u_1+u_2+...+u_n\right)=n\left[2u_1+\left(n-1\right)d\right]\\ \Leftrightarrow2\left(u_1+u_2+...+u_n\right)=n\left(u_1+u_n\right)\)
Gọi số hạng đầu và công bội của cấp số nhân là: \(u_1;q\).
a) Theo tính chất của cấp số nhân ta có:
\(\left\{{}\begin{matrix}u_1q^4-u_1=15\\u_1q^3-u_1q=6\end{matrix}\right.\)\(\Rightarrow\dfrac{u_1\left(q^4-1\right)}{u_1\left(q^3-q\right)}=\dfrac{15}{6}\)\(\Leftrightarrow\dfrac{\left(q^2-1\right)\left(q^2+1\right)}{q\left(q^2-1\right)}=\dfrac{15}{6}\)\(\Leftrightarrow\dfrac{q^2+1}{q}=\dfrac{15}{6}\)
\(\Leftrightarrow6\left(q^2+1\right)=15q\)\(\Leftrightarrow6q^2-15q+6=0\)\(\Leftrightarrow\left[{}\begin{matrix}q=2\\q=\dfrac{1}{2}\end{matrix}\right.\).
Với \(q=2\).
Suy ra: \(u_1\left(q^4-q\right)=15\Rightarrow u_1=\dfrac{15}{q^4-q}=\dfrac{15}{14}\).
Với \(q=\dfrac{1}{2}\)
Suy ra \(u_1=\dfrac{15}{q^4-q}=\dfrac{-240}{7}\).
a) \(\left\{ \begin{array}{l}{u_5} = 96\\{u_6} = 192\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u_1}.{q^4} = 96\\{u_1}.{q^5} = 192\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u_1}.{q^4} = 96\\\left( {{u_1}.{q^4}} \right).q = 192\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u_1}.{q^4} = 96\\96q = 192\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}q = 2\\{u_1} = 6\end{array} \right.\)
Vậy cấp số nhân \(\left( {{u_n}} \right)\) có số hạng đầu \({u_1} = 6\) và công bội \(q = 2\).
b)
\(\left\{ \begin{array}{l}{u_4} + {u_2} = 60\\{u_5} - {u_3} = 144\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u_1}.{q^3} + {u_1}.q = 60\\{u_1}.{q^4} - {u_1}.{q^2} = 144\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u_1}.q\left( {{q^2} + 1} \right) = 60\left( 1 \right)\\{u_1}.{q^2}\left( {{q^2} - 1} \right) = 144\left( 2 \right)\end{array} \right.\)
Do \({u_1} = 0\) và \(q = 0\) không là nghiệm của hệ phương trình nên chia vế với vế của (2) cho (1) ta được:
\(\frac{{q\left( {{q^2} - 1} \right)}}{{{q^2} + 1}} = \frac{{144}}{{60}} \Leftrightarrow \frac{{q\left( {{q^2} - 1} \right)}}{{{q^2} + 1}} =\frac{{12}}{{5}} \Leftrightarrow 5q\left( {{q^2} - 1} \right) = 12\left( {{q^2} + 1} \right)\)
\( \Leftrightarrow 5{q^3} - 12q = 5{q^2} + 12 \Leftrightarrow 5{q^3} - 12{q^2} - 5q - 12 = 0 \Leftrightarrow q=3\) thế vào (1) ta được \({u_1}=2\).
Vậy cấp số nhân \(\left( {{u_n}} \right)\) có số hạng đầu \({u_1} = 2\) và công bội \(q = 3\).
Ta có:
\(u_2=u_1.q\\ u_3=u_2.q=\left(u_1.q\right).q=u_1.q^2\\ u_4=u_3.q=\left(u_1.q^2\right).q=u_1.q^3.\\ .\\ .\\ .\\ u_{10}=u_1.q^9\)