Showing that sum, $ \small 2^k - 2^{k-1} + 2^{k-2} + ..... + (-1)^k 2^0 = \frac{2^{k+1} + (-1)^k}{3} $

Calculation:

Let, $ \large S = 2^k - 2^{k-1} + 2^{k-2} - 2^{k-3} + ..... + (-1)^k 2^0 \qquad \qquad (i) $

This can also be calculated using the series below,
$ \large S = 2^k - \frac{2^k}{2} + \frac{2^k}{4} - \frac{2^k}{8} + ..... + (-1)^k \frac{2^k}{2^k} $

$ \large S = 2^k( 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + ..... + (-1)^k \frac{1}{2^k} ) $

But direct calculation is faster. Notice that dividing a term by $ 2 $ gives the next term.
$ \large \frac{S}{2} = \frac{2^k}{2} - \frac{2^{k-1}}{2} + \frac{2^{k-2}}{2} + ..... + (-1)^k \frac{2^0}{2} $

$ \large \frac{S}{2} = 2^{k-1} - 2^{k-2} + 2^{k-3} + ..... + (-1)^k 2^{-1} \qquad \qquad (ii) $

Now adding $ (i) $ and $ (ii) $ will cancel the internal terms,
$ \large S + \frac{S}{2} = 2^k + (-1)^k 2^{-1} $

$ \large \frac{3S}{2} = 2^k + (-1)^k 2^{-1} $

$ \large 3S = 2^{k+1} + (-1)^k \frac{2}{2} $

$ \large S = \frac{2^{k+1} + (-1)^k}{3} $

Finally setting the value of $ S $,
$ \large 2^k - 2^{k-1} + 2^{k-2} + ..... + (-1)^k 2^0 = \frac{2^{k+1} + (-1)^k}{3} $