\(\Lambda_{\mathrm{n}}=\frac{1}{R Z^{2}}\left(1-\frac{1}{\mathrm{n}^{2}}\right)^{-1}\)
since \(n\) is very large, using binomial theorem
\(\Lambda_{n}=\frac{1}{R Z^{2}}\left(1+\frac{1}{n^{2}}\right)\)
\(\Lambda_{\mathrm{n}}=\frac{1}{R Z^{2}}+\frac{1}{R Z^{2}}\left(\frac{1}{\mathrm{n}^{2}}\right)\)
As we know, \(\lambda_{\mathrm{n}}=\frac{2 \pi \mathrm{r}}{\mathrm{n}}=2 \pi\left(\frac{\mathrm{n}^{2} \mathrm{h}^{2}}{4 \pi^{2} \mathrm{m} \mathrm{Ze}^{2}}\right) \frac{1}{\mathrm{n}} \propto \mathrm{n}\)
\(\Lambda_{\mathrm{n}} \approx \mathrm{A}+\frac{\mathrm{B}}{\lambda_{\mathrm{n}}^{2}}\)