Let f: R → R satisfy f(x+y) = f(x)f(y) for all x, y ∈ R and f(1) = 2 — JEE Mathematics

The functional equation $f(x+y) = f(x)f(y)$ represents an exponential function of the form $f(x) = a^x$.
Given $f(1) = 2$, we have $a^1 = 2 \implies a = 2$.
So, $f(x) = 2^x$.

Now evaluate the summation:

$$\sum_{k=1}^{n} f(k) = \sum_{k=1}^{n} 2^k = 2^1 + 2^2 + 2^3 + \dots + 2^n$$

This is a geometric progression with first term $A = 2$, common ratio $r = 2$, and $n$ terms.

$$Sum = A \frac{r^n - 1}{r - 1} = 2 \frac{2^n - 1}{2 - 1} = 2(2^n - 1)$$

Answer: $2(2^n - 1)$