ข้อสอบ PAT 1 - กุมภาพันธ์ 2562

$$\begin{gathered} \mathrm{จากโจทย์} f\left(n\right) = f\left(n-2\right)+2n \to \boxed{1} \\ \mathrm{เมื่อ} n \mathrm{เป็นจำนวนคี่} \\ \mathrm{จะได้} f\left(5\right) = f\left(5-2\right)+2\left(5\right) \\ f\left(5\right) = f\left(3\right)+10 ; \mathrm{โดย} f\left(5\right) =16 \\ 16 = f\left(3\right)+10 \\ f\left(3\right) = 6 \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{1} f\left(3\right) = f\left(3-2\right)+2\left(3\right) \\ f\left(3\right) = f\left(1\right)+6 \\ 6 = f\left(1\right)+6 \\ f\left(1\right) = 0 \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{1} f\left(1\right) = f\left(1-2\right)+2\left(1\right) \\ 0 = f\left(-1\right)+2 \\ f\left(-1\right) = -2 \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{1} f\left(-1\right) = f\left(-1-2\right)+2\left(-1\right) \\ -2 = f\left(-3\right)-2 \\ f\left(-3\right) = 0 \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} f\left(n\right) = f\left(n+1\right)-n \to \boxed{2} \\ \mathrm{เมื่อ} n \mathrm{เป็นจำนวนคู่} \\ \mathrm{จะได้} f\left(2\right) = f\left(2+1\right)-2 \\ f\left(2\right) = f\left(3\right)-2 ; \mathrm{โดย} f\left(3\right)=6 \\ f\left(2\right) = 6-2 \\ f\left(2\right) = 4 \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{2} f\left(0\right) = f\left(0+1\right)-0 \\ f\left(0\right) = f\left(1\right) ; \mathrm{โดย} f\left(1\right)=0 \\ f\left(0\right) = 0 \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{2} f\left(-2\right) = f\left(-2+1\right)-\left(-2\right) \\ f\left(-2\right) = f\left(-1\right)+2 ; \mathrm{โดย} f\left(-1\right)=-2 \\ f\left(-2\right) = -2+2 \\ f\left(-2\right) = 0 \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \sum _{n=-3}^{3}f\left(n\right) = f\left(-3\right)+f\left(-2\right)+f\left(-1\right)+...+f\left(3\right) \\ = 0+0+\left(-2\right)+0+0+4+6 \\ = 8 \end{gathered}$$