\(U_n=\dfrac{\left(n^2-1\right)}{n\left(n+2\right)}U_{n-1}\Rightarrow n\left(n+2\right).U_n=\left(n-1\right)\left(n+1\right).U_{n-1}\)

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 \(V_1=1.\left(1+2\right).U_1=1\)

\(\Rightarrow V_n=1\)

\(\Rightarrow U_n=\dfrac{V_n}{n\left(n+2\right)}=\dfrac{1}{n\left(n+2\right)}\)

\(\Rightarrow A=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{2015.2017}\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\right)\)

\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)\)

\(=...\)

\(\dfrac{u_{n+1}}{n+1}=3.\dfrac{u_n}{n}\)

Đặt \(\dfrac{u_n}{n}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=\dfrac{1}{3}\\v_{n+1}=3v_n\end{matrix}\right.\)

\(\Rightarrow v_n=\dfrac{1}{3}.3^{n-1}=3^{n-2}\)

\(\Rightarrow S=3^{-1}+3^0+...+3^8=...\)

Bạn tham khảo câu trả lời của anh Lâm

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\(u_{n+1}=\dfrac{u_n}{u_n+1}\Rightarrow\dfrac{1}{u_{n+1}}=\dfrac{1}{u_n}+1\)

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 \(d=1\Rightarrow v_n=v_1+\left(n-1\right).1=n\)

\(\Rightarrow u_n=\dfrac{1}{n}\)

\(\Rightarrow u_n+1=\dfrac{n+1}{n}\)

\(\lim\dfrac{2014\left(\dfrac{2}{1}\right)\left(\dfrac{3}{2}\right)\left(\dfrac{4}{3}\right)...\left(\dfrac{n+1}{n}\right)}{2015n}=\lim\dfrac{2014\left(n+1\right)}{2015n}=\dfrac{2014}{2015}\)

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: \(S_{10}=u_1+u_2+...+u_9+u_{10}\)

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

\(=-6\cdot\left(1-\dfrac{1}{2^{10}}\right)=-6+\dfrac{6}{2^{10}}=-\dfrac{3069}{512}\)