ข้อสอบคณิต 9 สามัญ - ปี 2559

 $$\begin{gathered} \mathrm{จากโจทย์} {\mathrm{a}}_{\mathrm{n}}={\int }_{\mathrm{n}}^{\frac{\mathrm{n}\left(\mathrm{n}+2\right)}{2}}\frac{1}{{\mathrm{x}}^2}d\mathrm{x} \\ \mathrm{จะได้} {\mathrm{a}}_{\mathrm{n}}={\overline{)\frac{{\mathrm{x}}^{-1}}{-1}}}_{\mathrm{n}}^{\frac{\mathrm{n}\left(\mathrm{n}+2\right)}{2}} \\ ={\overline{)-{\mathrm{x}}^{-1}}}_{\mathrm{n}}^{\frac{\mathrm{n}\left(\mathrm{n}+2\right)}{2}} \\ =-\left({\left(\frac{\mathrm{n}\left(\mathrm{n}+2\right)}{2}\right)}^{-1}-{\left(n\right)}^{-1}\right) \end{gathered}$$

$$\begin{gathered} =-\left(\frac{2}{\mathrm{n}\left(\mathrm{n}+2\right)}-\frac{1}{\mathrm{n}}\right) \\ =-\left(\frac{2-\mathrm{n}-2}{\mathrm{n}\left(\mathrm{n}+2\right)}\right) \\ =\frac{\mathrm{n}}{\mathrm{n}\left(\mathrm{n}+2\right)} \\ =\frac{1}{\mathrm{n}+2} \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \sum _{\mathrm{n}=1}^{\infty }\frac{{\mathrm{a}}_{\mathrm{n}}}{\mathrm{n}}=\sum _{\mathrm{n}=1}^{\infty }\frac{1}{{\displaystyle \mathrm{n}\left(\mathrm{n}+2\right)}} \\ =\frac{1}{2}\left(\sum _{\mathrm{n}=1}^{\infty }\left(\frac{1}{{\displaystyle \mathrm{n}}}-\frac{1}{n+2}\right)\right) \\ =\frac{1}{2}\left((\frac{1}{1}-\cancel{\frac{1}{3}})+(\frac{1}{2}-\cancel{\frac{1}{4}})+(\cancel{\frac{1}{3}}-\cancel{\frac{1}{5}})+(\cancel{\frac{1}{4}}-\cancel{\frac{1}{6}})+...\right) \\ =\frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}\right) \\ =\frac{1}{2}\left(\frac{3}{2}\right) \\ =\frac{3}{4} \end{gathered}$$