ข้อสอบ PAT 1 - มีนาคม 2555

$$\begin{gathered} \mathrm{จากโจทย์} \hspace{0.17em}\sqrt{1+\frac{{\displaystyle 1}}{{\displaystyle {\mathrm{n}}^2}}+\frac{1}{(\mathrm{n}+1{)}^2}} \\ =\sqrt{\frac{{\mathrm{n}}^2(\mathrm{n}+1{)}^2+(\mathrm{n}+1{)}^2+{\mathrm{n}}^2}{{\mathrm{n}}^2(\mathrm{n}+1{)}^2}} \\ =\sqrt{\frac{{\mathrm{n}}^2(\mathrm{n}+1{)}^2+2{\mathrm{n}}^2+2\mathrm{n}+1}{{\left(\mathrm{n}(\mathrm{n}+1)\right)}^2}} \\ =\sqrt{\frac{{\left(\mathrm{n}(\mathrm{n}+1)\right)}^2+2\mathrm{n}\left(\mathrm{n}+1\right)+1}{{\left(\mathrm{n}(\mathrm{n}+1)\right)}^2}} \end{gathered}$$

$$\begin{gathered} =\sqrt{\frac{{\left(\mathrm{n}(\mathrm{n}+1)+1\right)}^2}{{\left(\mathrm{n}(\mathrm{n}+1)\right)}^2}} \\ =\frac{\mathrm{n}(\mathrm{n}+1)+1}{\mathrm{n}(\mathrm{n}+1)} \\ =1+\frac{1}{\mathrm{n}(\mathrm{n}+1)} \\ =1+\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1} \\ \mathrm{จะได้} \hspace{0.17em}\sqrt{1+\frac{{\displaystyle 1}}{{\displaystyle {\mathrm{n}}^2}}+\frac{1}{(\mathrm{n}+1{)}^2}}=1+\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1} \end{gathered}$$

$$\begin{gathered} \mathbf{หาค่าของ} \\ \sqrt{1+\frac{{\displaystyle 1}}{{\displaystyle 1^2}}+\frac{1}{{\displaystyle 2^2}}}+\sqrt{1+\frac{{\displaystyle 1}}{{\displaystyle 2^2}}+\frac{1}{3^2}}+...+\sqrt{1+\frac{{\displaystyle 1}}{{\displaystyle {\mathrm{n}}^2}}+\frac{1}{(\mathrm{n}+1{)}^2}} \\ =\left(1+\frac{{\displaystyle 1}}{{\displaystyle 1}}-\frac{1}{{\displaystyle 2}}\right)+\left(1+\frac{{\displaystyle 1}}{{\displaystyle 2}}-\frac{1}{{\displaystyle 3}}\right)+\left(1+\frac{{\displaystyle 1}}{{\displaystyle 3}}-\frac{1}{{\displaystyle 4}}\right)+...+\left(1+\frac{{\displaystyle 1}}{{\displaystyle \mathrm{n}}}+\frac{1}{\mathrm{n}+1}\right) \\ =\left(1+1+1+...+1\right)+\left[\left(\frac{{\displaystyle 1}}{{\displaystyle 1}}-\frac{1}{{\displaystyle 2}}\right)+\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}-\frac{1}{{\displaystyle 3}}\right)+\left(\frac{{\displaystyle 1}}{{\displaystyle 3}}-\frac{1}{{\displaystyle 4}}\right)+...+\left(\frac{{\displaystyle 1}}{{\displaystyle \mathrm{n}}}+\frac{1}{\mathrm{n}+1}\right)\right] \\ =\mathrm{n}+1-\frac{1}{\mathrm{n}+1} \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \\ \underset{\mathrm{n}\to \infty }{\lim }\frac{1}{\mathrm{n}}\left(\sqrt{1+\frac{{\displaystyle 1}}{{\displaystyle 1^2}}+\frac{1}{{\displaystyle 2^2}}}+\sqrt{1+\frac{{\displaystyle 1}}{{\displaystyle 2^2}}+\frac{1}{3^2}}+...+\sqrt{1+\frac{{\displaystyle 1}}{{\displaystyle {\mathrm{n}}^2}}+\frac{1}{(\mathrm{n}+1{)}^2}}\right) \end{gathered}$$

$$\begin{gathered} =\underset{\mathrm{n}\to \infty }{\lim }\frac{1}{\mathrm{n}}(\mathrm{n}+1-\frac{1}{\mathrm{n}+1}) \\ =\underset{\mathrm{n}\to \infty }{\lim }\frac{1}{\mathrm{n}}(\mathrm{n}+1) \\ =1 \end{gathered}$$