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$$\Delta_{E_{n}}=n(E_{a}+n\frac{\lambda}{L} E_{a})-(n-1)(E_{a}+(n-1)\frac{\lambda}{L} E_{a})$$  $$\Delta_{E_{n}}=E_{a}\frac{\lambda}{L} (2n-1)+E_{a}$$  Therefore, when n=$\frac{1}{2}$:  $$\Delta_{E_{n}}=E_{a}$$ In other words:  $$E_{a}=\Delta_{E(0.5)}$$