## Identity of two equal sums of n distinct squares

Establish the following identity:

$\sum_{k=0}^{n-2} (2^k - 1)^2 \; + \; (3(2^{n-1}) - 1)^2 \; = \; \sum_{k=2}^{n} (2^k + 1)^2 \; + \; (2^n - 4)^2$