Using the equation; $F_{r}=12\frac{U_{0}}{R_{0}}\left[\left(\frac{R_{0}}{R_{0}+x}\right)^{13}-\left(\frac{R_{0}}{R_{0}+x}\right)^{7}\right]$ I must apply the binomial theroem to; $(\frac{1}{R_{0}+x})^{13}$ And $(\frac{1}{R_{0}+x})^7$,
In order to show that $F_{r}\approx -\left(\frac{72U_{0}}{R_{0}^{2}}\right)x$
Also, x is very small comapared to $R_{0}$ so, $\left| \frac{x}{R_{0}}\right|<1$
What Ive done so far is, $\frac{1}{R^{13}_{0}}.\frac{1}{1+(\frac{x}{R_{0}})^{13}}$ $=\frac{1}{R^{13}_{0}}\left(1+(\frac{x}{R_{0}})\right)^{-13}\approx \frac{1}{R^{13}_{0}}\left(1+(-13)\frac{x}{R_{0}}\right)=\frac{1}{R^{13}_{0}}\left(1-13\frac{x}{R_{0}}\right)$
So, does? $F_{r}\approx -\left(\frac{72U_{0}}{R_{0}^{2}}\right)x=12\frac{U_{0}}{R_{0}}\left[\frac{1}{R^{13}_{0}}\left(1-13\frac{x}{R_{0}}\right)-\frac{1}{R^{7}_{0}}\left(1-7\frac{x}{R_{0}}\right)\right]$
If so, could someone show me some working because I cant quite get to the right answer. Any help will be greatly appreciated, thanks