ข้อสอบคณิต 9 สามัญ - ปี 2555

$$\begin{gathered} \mathbf{สูตรทฤษฎีบททวินาม}\mathbf{ } \\ - {\left(\mathrm{a}+\mathrm{b}\right)}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{n}}+\left(\begin{array}{c}\mathrm{n}\\ 1\end{array}\right){\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+\left(\begin{array}{c}\mathrm{n}\\ 2\end{array}\right){\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^2+...+{\mathrm{b}}^{\mathrm{n}} \\ - \mathrm{พจน์ที่} \mathrm{r}+1={\mathrm{T}}_{\mathrm{r}+1}=\left(\begin{array}{c}\mathrm{n}\\ \mathrm{r}\end{array}\right){\mathrm{a}}^{\mathrm{n}-\mathrm{r}}{\mathrm{b}}^{\mathrm{r}} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} {\left({\mathrm{x}}^2+\frac{2}{{\mathrm{x}}^3}\right)}^{10} \\ \mathrm{กำหนดให้} \mathrm{n}=10 , \mathrm{a}={\mathrm{x}}^2 , \mathrm{b}=\frac{2}{{\mathrm{x}}^3} \\ \mathrm{จะได้} {\mathrm{T}}_{\mathrm{r}+1}=\left(\begin{array}{c}10\\ \mathrm{r}\end{array}\right){\left({\mathrm{x}}^2\right)}^{10-\mathrm{r}}{\left(\frac{2}{{\mathrm{x}}^3}\right)}^{\mathrm{r}} \\ =\left(\begin{array}{c}10\\ \mathrm{r}\end{array}\right){\mathrm{x}}^{20-2\mathrm{r}}·\frac{2^{\mathrm{r}}}{{\mathrm{x}}^{3\mathrm{r}}} \\ =2^{\mathrm{r}}·\left(\begin{array}{c}10\\ \mathrm{r}\end{array}\right)·\frac{{\mathrm{x}}^{20-2\mathrm{r}}}{{\mathrm{x}}^{3\mathrm{r}}}\to \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} \mathrm{ในการกระจายจะได้ว่าพจน์ค่าคงตัว} \\ \mathrm{จาก} \boxed{1} {\mathrm{T}}_{\mathrm{r}+1}=2^{\mathrm{r}}·\left(\begin{array}{c}10\\ \mathrm{r}\end{array}\right)·\frac{{\mathrm{x}}^{20-2\mathrm{r}}}{{\mathrm{x}}^{3\mathrm{r}}} \\ \mathrm{แสดงว่า} 20-2\mathrm{r}=3\mathrm{r} \\ 20=5\mathrm{r} \\ \mathrm{r}=4 ; \mathrm{แทนใน} \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} {\mathrm{T}}_{\mathrm{r}+1}=2^4·\left(\begin{array}{c}10\\ 4\end{array}\right) \\ =16·\frac{10!}{6!·4!} \\ = 16·\frac{10\times 9\times 8\times 7\times \cancel{6!}}{\cancel{6!}·4!} \\ =16\left(210\right) \\ =3360 \\ \mathrm{ดังนั้น} \mathrm{พจน์ค่าคงตัวมีค่าเท่ากับ} 3360 \end{gathered}$$