\(\left\{{}\begin{matrix}u_1=1\\u_{n+1}=\dfrac{u_n^{2016}}{2015}+u_n\end{matrix}\right.\). Tính \(s=lim\left(\dfrac{u_1^{2015}}{u_2}+\dfrac{u_2^{2015}}{u_3}+...+\dfrac{u_n^{2015}}{u_{n+1}}\right)\)
Những câu hỏi liên quan
\(\left\{{}\begin{matrix}u_1=2\\u_{n+1}=\dfrac{u_n^2+2016u_n}{2017}\end{matrix}\right.\). Tính \(limS;S=\dfrac{u_1}{u_2-1}+\dfrac{u_2}{u_3-1}+...+\dfrac{u_n}{u_{n+1}-1}\)
\(\left\{{}\begin{matrix}u_1=1\\u_{n+1}=\dfrac{1}{3}\left(1+\dfrac{1}{u_n}\right)u_n\end{matrix}\right.\). gọi \(S_n=u_1+\dfrac{u_2}{2}+\dfrac{u_3}{3}+...+\dfrac{u_n}{n}\). tìm \(\lim\limits S_n\)
Cho dãy un được xác định bởi
\(\left\{{}\begin{matrix}u_1=1\\u_{n+1}=\dfrac{u_n}{u_n+1}\end{matrix}\right.\) với n=1,2,3,.... Tính
\(\lim\limits_{ }\dfrac{2014\left(u_1+1\right)\left(u_2+1\right)....\left(u_n+1\right)}{2015n}\)
\(u_{n+1}=\dfrac{u_n}{u_n+1}\Rightarrow\dfrac{1}{u_{n+1}}=\dfrac{1}{u_n}+1\)
Đặt \(\dfrac{1}{u_n}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=\dfrac{1}{u_1}=1\\v_{n+1}=v_n+1\end{matrix}\right.\)
\(\Rightarrow v_n\) là CSC với công sai \(d=1\Rightarrow v_n=v_1+\left(n-1\right).1=n\)
\(\Rightarrow u_n=\dfrac{1}{n}\)
\(\Rightarrow u_n+1=\dfrac{n+1}{n}\)
\(\lim\dfrac{2014\left(\dfrac{2}{1}\right)\left(\dfrac{3}{2}\right)\left(\dfrac{4}{3}\right)...\left(\dfrac{n+1}{n}\right)}{2015n}=\lim\dfrac{2014\left(n+1\right)}{2015n}=\dfrac{2014}{2015}\)
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\(\left\{{}\begin{matrix}u_1=\dfrac{1}{2};u_2=3\\u_{n+2}=\dfrac{u_{n+1}.u_n+1}{u_{n+1}+u_n}\end{matrix}\right.\). tìm \(\left(u_n\right)\)
cho dãy số (un):\(\left\{{}\begin{matrix}u_1=3\\u_{n+1}=u_n^2-3u_n+4\end{matrix}\right.\)
Tìm lim\(\left(\dfrac{1}{u_1-1}+\dfrac{1}{u_2-1}+...+\dfrac{1}{u_n-1}\right)\)
Bạn tham khảo câu trả lời của anh Lâm
https://hoc24.vn/cau-hoi/.334447965337
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6 tháng 1 lúc 11:58
Cho dãy số \(\left\{U_n\right\}\) được xác định như sau: \(U_1=\dfrac{1}{3},U_n=\dfrac{\left(n^2-1\right)U_{n-1}}{n\left(n+2\right)}\) (Với \(n=2;3;4...\)). Tính gần đúng giá trị của biểu thức \(A=U_1+U_2+U_3+...+U_{2015}\).
\(U_n=\dfrac{\left(n^2-1\right)}{n\left(n+2\right)}U_{n-1}\Rightarrow n\left(n+2\right).U_n=\left(n-1\right)\left(n+1\right).U_{n-1}\)
Đặt \(n\left(n+2\right).U_n=V_n\Rightarrow V_{n-1}=\left(n-1\right)\left(n+2-1\right).U_{n-1}=\left(n-1\right).\left(n+1\right)U_{n-1}\)
\(\Rightarrow V_n=V_{n-1}\)
\(\Rightarrow V_n=V_{n-1}=V_{n-2}=...=V_1\)
Có \(V_1=1.\left(1+2\right).U_1=1\)
\(\Rightarrow V_n=1\)
\(\Rightarrow U_n=\dfrac{V_n}{n\left(n+2\right)}=\dfrac{1}{n\left(n+2\right)}\)
\(\Rightarrow A=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{2015.2017}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\right)\)
\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)\)
\(=...\)
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Cho dãy (un) \(\left\{{}\begin{matrix}u_1=\dfrac{1}{2}\\u_n=\dfrac{\sqrt{u_{n-1}^2+4u_{n-1}}+u_{n-1}}{2}\forall n\ge2\end{matrix}\right.\)
Tinh \(\lim\limits_{n\rightarrow+\infty}\left(\dfrac{1}{u_1^2}+\dfrac{1}{u_2^2}+...+\dfrac{1}{u_n^2}\right)\)
Cho \(\left(U_n\right):\left\{{}\begin{matrix}u_1=2019\\u_n=\dfrac{-2019}{n}.\left(u_1+u_2+...+u_{n-1}\right)\end{matrix}\right.\). Tính: \(A=2u_1+2^2u_2+...+2^{2019}u_{2019}\)
Cho dãy \(u_n\) thỏa\(\left\{{}\begin{matrix}u_1=a,u_2=b\\u_{n+2}=\dfrac{u_{n+1}+u_n}{2}\end{matrix}\right.\). TÍnh \(limu_n\)
\(\left\{{}\begin{matrix}u_1=a;u_2=b\\u_{n+2}=\dfrac{1}{2}u_{n+1}+\dfrac{1}{2}u_n\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u_1=a,u_2=b\\u_{n+2}+\dfrac{1}{2}u_{n+1}=u_{n+1}+\dfrac{1}{2}u_n\end{matrix}\right.\)
\(v_{n+1}=u_{n+1}+\dfrac{1}{2}u_n\Rightarrow\left\{{}\begin{matrix}v_2=u_2+\dfrac{1}{2}u_1=b+\dfrac{1}{2}a\\v_{n+1}=v_n\end{matrix}\right.\)
\(\Rightarrow v_{n+1}=b+\dfrac{1}{2}a\Rightarrow u_{n+1}=b+\dfrac{1}{2}a-\dfrac{1}{2}u_n\)
\(\Leftrightarrow u_{n+1}-\left(\dfrac{1}{3}a+\dfrac{2}{3}b\right)=-\dfrac{1}{2}\left[u_n-\left(\dfrac{1}{3}a+\dfrac{2}{3}b\right)\right]\)
\(t_n=u_n-\left(\dfrac{1}{3}a+\dfrac{2}{3}b\right)\Rightarrow\left\{{}\begin{matrix}t_1=u_1-\dfrac{1}{3}a-\dfrac{2}{3}b=\dfrac{2}{3}\left(a-b\right)\\t_{n+1}=-\dfrac{1}{2}t_n\end{matrix}\right.\)
\(\Rightarrow t_n=\dfrac{2}{3}\left(a-b\right)\left(-\dfrac{1}{2}\right)^{n-1}\Rightarrow u_n=t_n+\dfrac{1}{3}a+\dfrac{2}{3}b=\dfrac{2}{3}\left(a-b\right)\left(-\dfrac{1}{2}\right)^{n-1}+\dfrac{1}{3}a+\dfrac{2}{3}b\)
\(\Rightarrow limun=\lim\limits\left[\dfrac{2}{3}\left(a-b\right)\left(-\dfrac{1}{2}\right)^{n-1}+\dfrac{1}{3}a+\dfrac{2}{3}b\right]=0\)
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