Quay lại đáp án

c) Với mọi số tự nhiên $k$ thì ta biết $\quad\dfrac{2}{k(k+2)}=\dfrac{1}{k}-\dfrac{1}{k+2}$Do đó $2u_n=\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{n(n+2)}=\left ( \dfrac{1}{1}-\dfrac{1}{3} \right )+\left ( \dfrac{1}{2}-\dfrac{1}{4} \right )+\ldots+\left ( \dfrac{1}{n}-\dfrac{1}{n+2} \right )$$2u_n=\left (1+\dfrac{1}{2}+...+\dfrac{1}{n} \right )-\left (\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n+2} \right )$$2u_n=\left (1+\dfrac{1}{2} \right )-\left (\dfrac{1}{n+1}+\dfrac{1}{n+2} \right )$ $2u_n=\dfrac{n (5+3 n)}{2 (1+n) (2+n)}$ $u_n=\dfrac{n (5+3 n)}{4 (1+n) (2+n)}$ Ta sẽ chứng minh $0 <u_n \le \dfrac{3}{4}$.Làm tương tự như câu a và b ta có được điều này.

c) Với mọi số tự nhiên $k$ thì ta biết $\quad\dfrac{2}{k(k+2)}=\dfrac{1}{k}-\dfrac{1}{k+2}$Do đó $2u_n=\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{n(n+2)}=\left ( \dfrac{1}{1}-\dfrac{1}{3} \right )+\left ( \dfrac{1}{2}-\dfrac{1}{4} \right )+\ldots+\left ( \dfrac{1}{n}-\dfrac{1}{n+2} \right )$$2u_n=\left (1+\dfrac{1}{2}+...+\dfrac{1}{n} \right )-\left (\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n+2} \right )$$2u_n=\left (1+\dfrac{1}{2} \right )-\left (\dfrac{1}{n+1}+\dfrac{1}{n+2} \right )$ $2u_n=\dfrac{n (5+3 n)}{2 (1+n) (2+n)}$ $u_n=\dfrac{n (5+3 n)}{4 (1+n) (2+n)}$ Ta sẽ chứng minh $\dfrac{1}{4} <u_n &lt; \dfrac{3}{4}$.Thật vậy$\dfrac{n (5+3 n)}{4 (1+n) (2+n)}< \dfrac{3}{4}\Leftrightarrow n (5+3 n) &lt;3(1+n) (2+n)$$\Leftrightarrow 3n^2+5n &lt; 3n^2+9n+6\Leftrightarrow 0 < 4n+6$, luôn đúng. $\dfrac{n (5+3 n)}{4 (1+n) (2+n)}>\dfrac{1}{4}\Leftrightarrow n (5+3 n) >(1+n) (2+n)$$\Leftrightarrow 3n^2+5n > n^2+3n+2\Leftrightarrow 2n^2+2n > 2$, luôn đúng.