Answer
$\Delta E = (\frac{h^2}{2m~L^2})~(n+1)$
Work Step by Step
We can find the energy difference between $E_{n+2}$ and $E_n$:
$E_{n+2} - E_n = (\frac{h^2}{8m~L^2})~(n+2)^2 - (\frac{h^2}{8m~L^2})~n^2$
$E_{n+2} - E_n = (\frac{h^2}{8m~L^2})~(n^2+4n+4) - (\frac{h^2}{8m~L^2})~n^2$
$E_{n+2} - E_n = (\frac{h^2}{8m~L^2})~(4n+4)$
$E_{n+2} - E_n = (\frac{4~h^2}{8m~L^2})~(n+1)$
$E_{n+2} - E_n = (\frac{h^2}{2m~L^2})~(n+1)$
Therefore:
$\Delta E = (\frac{h^2}{2m~L^2})~(n+1)$
