ข้อสอบคณิต O-NET - 2559

$$\begin{gathered} \mathrm{จากสูตร} \mathrm{อนุกรมเรขาคณิต} \\ S_n = \frac{a_1-a_{n}r}{1-r} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} 729A_6 = 729\left[a_2+a_3+a_4+a_5+a_6\right] \\ = 3^6\left[{\left(2\right)}^{2-2}{\left(\frac{1}{3}\right)}^2+{\left(2\right)}^{3-2}{\left(\frac{1}{3}\right)}^3+\dots +{\left(2\right)}^{6-2}{\left(\frac{1}{3}\right)}^6\right] \\ = 3^6\left(\frac{2^0}{3^2}+\frac{2^1}{3^3}+\frac{2^2}{3^4}+\frac{2^3}{3^5}+\frac{2^4}{3^6}\right) \\ = 3^4+2^1{3}^3+2^2{3}^2+2^3{3}^1+2^4 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} a_1=3^4 , r=\frac{2}{3} , a_n=2^4 \\ \mathrm{ดังนั้น} \mathrm{ผลบวก} = \frac{3^4-2^4\left({\displaystyle \frac{2}{3}}\right)}{1-{\displaystyle \frac{2}{3}}} \\ = \left(3^4-\frac{2^5}{3}\right)\left(\frac{3}{1}\right) \\ = 3^5-2^5 \\ = 243-32 \\ = 211 \to \mathrm{ตอบข้อ} 4 \end{gathered}$$