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$$\Delta E_{n}=E_{n}-E_{n-1}$$  $$\Delta E_{n}=n(E_{a}+\delta_{n})-(n-1)(E_{a}+\delta_{n-1})$$  $$\Delta E_{n}=n(E_{a}+n\frac{\lambda}{L} E_{a})-(n-1)(E_{a}+(n-1)\frac{\lambda}{L} E_{a})$$  $$\Delta E_{n}=E_{a}\frac{\lambda}{L} (2n-1)+E_{a}$$ (*) (2n-1)+E_{a} (*)$$  Therefore, when n=$\frac{1}{2}$:  $$\Delta E_{n}=E_{a}$$  In other words: