Answer
$|x_{n+1}-r|\leq \dfrac{M}{2K}|x_n-r|^2$
Work Step by Step
Consider $f(r)=f(x_n) +f'(x_n)(r-x_n)+R_1(r)$
For $f(r)=0$, we have $f(x_n) +f'(x_n)(r-x_n)+R_1(r)=0$
Re-arrange as:
$\dfrac{R_1(r)}{f'(x_n)}=x_n-r-\dfrac{f(x_n)}{f'(x_n)}(r-x_n)$
This implies that
$|x_{n+1}-r|=|\dfrac{R_1(r)}{f'(x_n)}|$
Apply Newton's formula as given in the statement.
$|R_1{r}|\leq |\dfrac{f''(r)}{2!}||r-x_n|^2$
Hence, we have $|x_{n+1}-r|\leq \dfrac{M}{2K}|x_n-r|^2$
