A natural functional equation

$f(x + y) = f(x).f(y)$

$\displaystyle f(a)\sum_{k = 1}^{n} f(k) = 16(\dfrac{2^{n} - 1}{2 - 1}$

$\displaystyle f(a)\sum_{k = 1}^{n} f(k) = \dfrac{16}{2}( 2 + 2^{2} + 2^{3} + 2^{4} + ..+ 2^{n}$

Since $f(1) = 2^{1}$ , by comparing $f(n) = 2^{n }$ , Thus $f(a) = 8 , a = 3$

i just put n = 1 and got that f(a+1)=16 which means f(a)f(1)=16 which means f(a)=8 and since i know that f(1)=2 and f(2)=4 then f(a)=f(1)*f(2) which means that a=1+2=3

seriesSum(f(a+k),k,1,n) = 16(2^n -1) ie seriesSum(2^(a+k),k,1,n) = 2^(a+1) (2^n -1) Thus, on comparing, 2^(a+1) = 16. Thus, a=3. f(x+y)=f(x)f(y) also f(1)=2. Thus, f(a)= f(1+a-1) = f(1)f(a-1) = 2f(a-1) = 2f(1+a-2) = 4f(a-2). Thus, f(a) = 2^r f(a-r). Thus, if r=a-1, then f(a)=2^a. Thus, similarly, f(a+k) = 2^(a+k).

This functional equation is pretty well known. I just realised that it had to be an exponential as it turned a sum f(x+y) into a product f(x)f(y)

From the given properties of $f$ , we can see:

$f(a+k) = 2^{a+k}$ , so we get

$\displaystyle \sum_{k=1}^n f(a+k) = 16(2^n-1) = \displaystyle \sum_{k=1}^n 2^{a+k} = 2^a \displaystyle \sum_{k=1}^n 2^k$ .

Using the formula for a geometric progression with $k$ not starting from $0$ and the ratio between consecutive terms not equal to $1$ , we get,

$2^a \displaystyle \sum_{k=1}^n 2^k = 2^a \frac{2^{n+1} - 2^1}{2^1 - 1} = 2^{a+1}(2^n - 1) = 16(2^n-1) \Rightarrow a = 3$ .