a

$(\begin{matrix} 10 \\ k + 1 \end{matrix}) = \frac{10!}{(k + 1)! (9 - k)!} = \frac{10! (10 - k)}{(k + 1) k! (10 - k) (9 - k)!} = \frac{10 - k}{k + 1} (\begin{matrix} 10 \\ k \end{matrix})$

b

$(\begin{matrix} 10 \\ k \end{matrix}) + (\begin{matrix} 10 \\ k + 1 \end{matrix}) = (\frac{10 - k}{k + 1} + 1) (\begin{matrix} 10 \\ k \end{matrix}) = \frac{11}{k + 1} \cdot (\begin{matrix} 10 \\ k \end{matrix})$

c

$Δ x = a$ en $Δ y = (\begin{matrix} 10 \\ k + 1 \end{matrix}) \cdot {(\frac{1}{2})}^{10} - (\begin{matrix} 10 \\ k \end{matrix}) \cdot {(\frac{1}{2})}^{10} =$ $(\begin{matrix} 10 \\ k \end{matrix}) \cdot {(\frac{1}{2})}^{10} \cdot (\frac{10 - k}{k + 1} - 1) =$ $(\begin{matrix} 10 \\ k \end{matrix}) \cdot {(\frac{1}{2})}^{10} \cdot \frac{9 - 2 k}{k + 1}$ , dus $\frac{Δ y}{Δ x} = \frac{1}{a} \cdot \frac{9 - 2 k}{k + 1} \cdot (\begin{matrix} 10 \\ k \end{matrix}) \cdot {(\frac{1}{2})}^{10}$ $\frac{Δ y}{Δ x} = \frac{1}{a} \cdot \frac{9 - 2 k}{k + 1} \cdot (\begin{matrix} 10 \\ k \end{matrix}) \cdot {(\frac{1}{2})}^{10}$

d

Uit a volgt: $9 - 2 k = ‐ \frac{2 x}{a}$ , dus $\frac{Δ y}{Δ x} = \frac{1}{a} \cdot \frac{9 - 2 k}{k + 1} \cdot (\begin{matrix} 10 \\ k \end{matrix}) \cdot {(\frac{1}{2})}^{10} =$
$\frac{1}{a} \cdot (9 - 2 k) \cdot \frac{1}{11} \cdot \frac{11}{k + 1} \cdot (\begin{matrix} 10 \\ k \end{matrix}) \cdot {(\frac{1}{2})}^{10} =$ $\frac{1}{a} \cdot ‐ \frac{2 x}{a} \cdot \frac{1}{11} \cdot 2 y =$ $‐ \frac{4}{11 a^{2}} \cdot x \cdot y$