Answer
$|x_{n+1}-r|\leq \dfrac{M}{2K}|x_n-r|^2$
Work Step by Step
Consider $f(r)=f(x_n) +f'(x_n)(r-x_n)+R_1(r)$
Let us suppose that $f(r)=0$
$f(r)=f(x_n) +f'(x_n)(r-x_n)+R_1(r) \implies f(x_n) +f'(x_n)(r-x_n)+R_1(r)=0$
This gives: $x_n-r-\dfrac{f(x_n)}{f'(x_n)}(r-x_n)=\dfrac{R_1(r)}{f'(x_n)}$
$\implies |x_{n+1}-r|=|\dfrac{R_1(r)}{f'(x_n)}| ...(a)$
Need to use Newton's formula.
$|R_1{r}|\leq |\dfrac{f''(r)}{2!}||r-x_n|^2$
Thus, the equation (a) can be arranged as: $|x_{n+1}-r|\leq \dfrac{M}{2K}|x_n-r|^2$
