ข้อสอบคณิต 9 สามัญ - ปี 2557

$$\begin{gathered} \mathrm{ทฤษฎีบททวินาม} {\left(a+b\right)}^n = \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)·a^{n-r}·b^r \\ \mathrm{จากโจทย์} \sum _{r=0}^n{\left(-1\right)}^r\left(\begin{array}{c}6\\ r\end{array}\right)·7^{6-r}·5^r \\ \mathrm{จะได้ว่า} n=6 , a=7 , b=-5 \end{gathered}$$

$$\begin{gathered} \mathrm{แสดงว่า} {\left[7+\left(-5\right)\right]}^6 = \sum _{r=0}^6\left(\begin{array}{c}6\\ r\end{array}\right)·7^{6-r}·{\left(-5\right)}^r \\ = \sum _{r=0}^6\left(\begin{array}{c}6\\ r\end{array}\right)·7^{6-r}·{\left(-1\right)}^r·5^r \\ = \sum _{r=0}^6{\left(-1\right)}^r\left(\begin{array}{c}6\\ r\end{array}\right)·7^{6-r}·5^r \\ \mathrm{ดังนั้น} \sum _{r=0}^6{\left(-1\right)}^r\left(\begin{array}{c}6\\ r\end{array}\right)·7^{6-r}·5^r = {\left[7+\left(-5\right)\right]}^6 \\ = {\left(7-5\right)}^6 = 2^6 \\ = 64 \end{gathered}$$