$$H_{1}=\frac{e^{2}}{R}+\frac{e^{2}}{R+x_{1}+x_{2}}-\frac{e^{2}}{R+x_{1}}-\frac{e^{2}}{R+x_{2}}$$ in the approximation $ \left |x_{1}\right |,\left |x_{2}\right |\ll R $ we expand to obtain in lowest order $$ H_{1} \simeq \frac{2e^{2}x_{1}x_{2}}{R^{3}} $$

Introduction to Solid State Physics,Charles Kittel, 8va edition,pag 55


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  • $\begingroup$ Hi and welcome to Physics Stack Exchange! I have a couple of comments though: firstly, it is preferred that formulas be written out in Mathjax code, rather than posted as images. Mathjax is explained on this page: math.meta.stackexchange.com/questions/5020/…. It can be embedded in questions and answers and makes them easier to read and more easily searchable. Secondly, you should include an image of the figure you are referencing. Also, it seems that you didn't complete the question? $\endgroup$ – Time4Tea Sep 26 '18 at 20:54

Use the approximation that $$1/(1-x)\simeq1+x+x^2$$ for small $x$. To get you started, I'll do the first two terms. Let us define $$H' =\frac{e^2}{R}+\frac{e^2}{R+(x_1-x_2)}.$$ Now define $z\equiv(x_2-x_1)/R$ and we have that $|z|\ll1$. We find then that $$\begin{align} H' &\equiv\frac{e^2}{R}\bigg(1+\frac{1}{1-z}\bigg) \\ &\simeq \frac{e^2}{R}\bigg(1+1+z+z^2+\cdots\bigg). \end{align}$$ Try doing the same for the remaining two terms. You should find that expanding to first order is insufficient due to cancellations, therefore a second order expansion is required. Let me know if you get stuck.


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