Q. Find $a$ for which $\sum\limits_{k=1}^{n}f\left(a+k\right) = 16\left(2^{n} - 1\right)$, where the function $f$ satisfies $f\left(x + y\right) =f\left(x\right)\cdot f\left(y\right)$ for all natural numbers $x, y$ and further $f\left(1\right) = 2$.
Sequences and Series
Solution: