Answer
$|x_{n+1}-r|\leq \dfrac{M}{2K}|x_n-r|^2$
Work Step by Step
Here, we have $f(r)=f(x_n) +f'(x_n)(r-x_n)+R_1(r)$
when $f(r)=0$
Then $f(x_n) +f'(x_n)(r-x_n)+R_1(r)=0$
or, $x_n-r-\dfrac{f(x_n)}{f'(x_n)}(r-x_n)=\dfrac{R_1(r)}{f'(x_n)}$
or, $|x_{n+1}-r|=|\dfrac{R_1(r)}{f'(x_n)}|$ ...(1)
According to Newton's formula, we have
$|R_1{r}|\leq |\dfrac{f''(r)}{2!}||r-x_n|^2$
The equation (1), becomes: $|x_{n+1}-r|\leq \dfrac{M}{2K}|x_n-r|^2$
