I'm having trouble with understanding where to begin, especially since this is my first time with recursions. From my understanding you'd have to match the both equations and simplify until they're equal with each other, or try to apply 2 into the equation of 1? Any hints or advice would be much appreciated!
2 is true for all real numbers r except r = 1 and for all integers n ≥ 0.
One can notice that $$2^n − 2^{n−1} + 2^{n−2} − 2^{n−3}+\cdots+ (−1)^{n−1}\cdot 2 + (−1)^n$$ can be seen as: \begin{align} S &= 2^n − 2^{n−1} + 2^{n−2} − 2^{n−3}+\cdots+ (−1)^{n−1}\cdot 2 + (−1)^n \\ &= 2^n + (-1)^1 \, 2^{n-1} + (-1)^2 \, 2^{n-2} + \cdots + (-1)^{n-1} \, 2^{n - (n-1)} + (-1)^n \, 2^{n - n} \\ &= \sum_{j=0}^{n} (-1)^j \, 2^{n-j} \\ &= 2^{n} \, \sum_{j=0}^{n} \left(- \frac{1}{2}\right)^j = 2^{n} \, \frac{1 - (-1/2)^{n+1}}{1 + \frac{1}{2}} = \frac{1}{3} \, (2^{n+1} - (-1)^{n+1}) \\ &= J_{n+1}. \end{align}
Note that: \begin{align} \sum_{j=0}^{n} t^j &= \frac{1 - t^{n+1}}{1-t} \\ J_{n} &= \frac{2^n - (-1)^n}{3} \hspace{5mm} \text{Jacobsthal numbers}. \end{align}
Further information about the Jacobsthal numbers can be found at Oeis A001045.