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\(u_1=5=5.1\)
\(u_2=10=5.2\)
\(u_3=15=5.3\)
....
\(\Rightarrow u_n=5n\)
\(u_{n+1}=\dfrac{3}{2}\left(u_n-\dfrac{n+4}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{3}{2}\left(u_n-\dfrac{3}{n+1}+\dfrac{2}{n+2}\right)\)
\(\Leftrightarrow u_{n+1}-\dfrac{3}{n+1+1}=\dfrac{3}{2}\left(u_n-\dfrac{3}{n+1}\right)\)
Đặt \(u_n-\dfrac{3}{n+1}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=u_1-\dfrac{3}{2}=-\dfrac{1}{2}\\v_{n+1}=\dfrac{3}{2}v_n\end{matrix}\right.\)
\(\Rightarrow v_n\) là CSN với công bội \(\dfrac{3}{2}\)
\(\Rightarrow v_n=-\dfrac{1}{2}\left(\dfrac{3}{2}\right)^{n-1}\)
\(\Rightarrow u_n=-\dfrac{1}{2}\left(\dfrac{3}{2}\right)^{n-1}+\dfrac{3}{n+1}\)
Từ công thức dãy số ta thấy \(u_n\) là cấp số cộng với \(\left\{{}\begin{matrix}u_1=2\\d=3\end{matrix}\right.\)
\(\Rightarrow u_n=u_1+\left(n-1\right)d=2+\left(n-1\right)3=3n-1\)
\(\Rightarrow I=\lim\limits\dfrac{3n-1}{3n+1}=1\)
\(\dfrac{1}{u_n-1}=\dfrac{1}{\dfrac{2^n-5^n}{2^n+5^n}-1}=\dfrac{2^n+5^n}{-2.5^n}=-\dfrac{1}{2}\left[\left(\dfrac{2}{5}\right)^n+1\right]\)
\(\Rightarrow S_n=-\dfrac{1}{2}\left[\left(\dfrac{2}{5}\right)^1+\left(\dfrac{2}{5}\right)^2+...+\left(\dfrac{2}{5}\right)^n+n\right]\)
Lại có: \(\left(\dfrac{2}{5}\right)^1+\left(\dfrac{2}{5}\right)^2+...+\left(\dfrac{2}{5}\right)^n=\dfrac{2}{5}.\dfrac{1-\left(\dfrac{2}{5}\right)^n}{1-\dfrac{2}{5}}=\dfrac{2}{3}\left[1-\left(\dfrac{2}{5}\right)^n\right]\)
\(\Rightarrow S_n=-\dfrac{1}{2}\left[\dfrac{2}{3}-\dfrac{2}{3}\left(\dfrac{2}{5}\right)^n+n\right]=...\)
cho dãy un \(\left\{{}\begin{matrix}U1=2\\Un+1=Un.\dfrac{n+1}{n}\end{matrix}\right.\)
tìm cttq dãy số
\(u_{n+1}=\dfrac{u_n\left(n+1\right)}{n}\Rightarrow\dfrac{u_{n+1}}{n+1}=\dfrac{u_n}{n}\)
Đặt \(\dfrac{u_n}{n}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=\dfrac{u_1}{1}=2\\v_{n+1}=v_n\end{matrix}\right.\)
\(\Rightarrow v_{n+1}=v_n=v_{n-1}=...=v_1=2\)
\(\Rightarrow\dfrac{u_n}{n}=2\Rightarrow u_n=2n\)
\(\Leftrightarrow\dfrac{u_{n+1}}{n+1}=\dfrac{1}{3}.\dfrac{u_n}{n}\)
Đặt \(\dfrac{u_n}{n}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=\dfrac{1}{3}\\v_{n+1}=\dfrac{1}{3}v_n\end{matrix}\right.\)
\(\Rightarrow v_n\) là CSN với công bội \(\dfrac{1}{3}\)
\(\Rightarrow v_n=\dfrac{1}{3}.\left(\dfrac{1}{3}\right)^{n-1}=\left(\dfrac{1}{3}\right)^n\)
\(S=\sum\limits^{10}_{k=1}\left(\dfrac{1}{3}\right)^k=\dfrac{\dfrac{1}{3}\left(1-\dfrac{1}{3^{10}}\right)}{1-\dfrac{1}{3}}=\dfrac{1}{2}\left(1-\dfrac{1}{3^{10}}\right)\)
C
0<Un≤\(\dfrac{1}{2}\) :))