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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
=>\(u1=-\dfrac{60}{-3-27}=\dfrac{60}{30}=2\)
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
a) \(\left\{{}\begin{matrix}u_2-u_3+u_5=10\\u_4+u_6=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1+d-u_1-2d+u_1+4d=10\\u_1+3d+u_1+5d=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1+3d=10\\2u_1+8d=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1=1\\d=3\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}u_2-u_6+u_4=-7\\u_8-2u_7=2u_4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1+d-u_1-5d+u_1+3d=-7\\u_1+7d-2\left(u_1+6d\right)=2\left(u_1+3d\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1-d=-7\\-3u_1-11d=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1=\dfrac{-11}{2}\\d=\dfrac{3}{2}\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}u_7-u_3=8\\u_2.u_7=75\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1+6d-u_1-2d=8\\\left(u_1+d\right)\left(u_1+6d\right)=75\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4d=8\\\left(u_1+d\right)\left(u_1+6d\right)=75\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}d=2\\\left(u_1+2\right)\left(u_1+12\right)=75\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}d=2\\u_1^2+14u_1+24=75\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}d=2\\\left[{}\begin{matrix}u_1=3\\u_1=-17\end{matrix}\right.\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}u_5=96\\u_7=384\end{matrix}\right.\)
\(u^2_6=u_5.u_7=96.384=36864\)
\(\Leftrightarrow u_6=192\)
\(q=\dfrac{u_7}{u_6}=\dfrac{384}{192}=2\)
\(u_5=u_1.q^4\)
\(\Leftrightarrow u_1=\dfrac{u_5}{q^4}=\dfrac{96}{2^4}=6\)
b) \(\left\{{}\begin{matrix}u_4-u_2=25\\u_3-u_1=50\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u_1.q^3-u_1.q=25\\u_1.q^2-u_1=50\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u_1.q\left(q^2-1\right)=25\left(1\right)\\u_1.\left(q^2-1\right)=50\left(2\right)\end{matrix}\right.\)
\(\left(1\right):\left(2\right)\Leftrightarrow q=\dfrac{25}{50}=\dfrac{1}{2}\)
\(\left(2\right)\Leftrightarrow u_1=\dfrac{50}{q^2-1}=\dfrac{50}{\dfrac{1}{4}-1}=-\dfrac{200}{3}\)
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: u4=4 và u6=8
=>u1+3d=4 và u1+5d=8
=>-2d=-4 và u1+3d=4
=>d=2 và u1=4-3d=-2
b: u1-u3+u5=10 và u1+u6=17
=>u1-u1-2d+u1+4d=10 và u1+u1+5d=17
=>u1+2d=10 và 2u1+5d=17
=>u1=16 và d=-3
c: u1+u2=5 và u3*u5=91
=>u1+u1+d=5 và (u1+2d)(u1+4d)=91
=>2u1+d=5 và (u1+2d)(u1+4d)=91
=>d=5-2u1 và (u1+10-4u1)(u1+20-8u1)=91
=>d=5-2u1 và (-3u1+10)(-7u1+20)=91
(-3u1+10)(-7u1+20)=91
=>21u1^2-60u1-70u1+200=91
=>21u1^2-130u1+109=0
=>u1=1 hoặc u1=109/21
Khi u1=1 thì d=5-2u1=5-2=3
Khi u1=109/21 thì d=5-2u1=5-218/21=-113/21
Đặt \(u_n+\dfrac{5}{4}=v_n\)
\(GT\Rightarrow\left\{{}\begin{matrix}v_1=\dfrac{9}{4};v_2=\dfrac{13}{4}\\v_{n+2}=2v_{n+1}+3v_n\end{matrix}\right.\)
Ta có CTTQ của dãy \(\left(v_n\right)\) là:
\(v_n=\dfrac{11}{24}.3^n-\dfrac{7}{8}.\left(-1\right)^n\)
(Bạn tự chứng minh theo quy nạp)
\(\Rightarrow u_n=\dfrac{11}{24}.3^n-\dfrac{7}{8}\left(-1\right)^n-\dfrac{5}{4}\) với \(\forall n\in N\text{*}\)
\(\Rightarrow S=2\left(u_1+u_2+...+u_{100}\right)+u_{101}\)
\(=\left[\dfrac{11}{12}\left(3^1+3^2+...+3^{100}\right)-\dfrac{7}{4}\left(-1+1-...+1\right)-\dfrac{5}{2}.100\right]+\dfrac{11}{24}.3^{101}-\dfrac{7}{8}.\left(-1\right)^{101}-\dfrac{5}{4}\)
\(=\dfrac{11}{12}.\dfrac{3^{101}-3}{2}-250+\dfrac{11}{24}.3^{101}+\dfrac{7}{8}\)
\(=\dfrac{11}{24}.\left(2.3^{101}-3\right)-\dfrac{1993}{8}\)
\(=\dfrac{11}{4}.3^{100}-\dfrac{501}{2}\)