Find the natural number ' $a$ ' for which $\sum_{k=1}^{n} f(a+k)=16\left(2^{n}-1\right)$ where the f

Question

Find the natural number ' $a$ ' for which $\sum_{k=1}^{n} f(a+k)=16\left(2^{n}-1\right)$ where the function ' $f$ ' satisfies the relation $f(x+y)=f(x) f(y)$ for all natural numbers $x, y$ and further $f(1)=2$.

Professor PC Thomas Classes · Answer

Given $f(x+y)=\mathrm{f}(x) f(y)$ for all natural numbers and $f(1)=2$\ $\Rightarrow \quad f(1+1)=f(1) \cdot f(1)=2^{2}$.\ We can prove by induction that\ $f(n)=2^{n}$ for all $n$\ Now by hypothesis\ $\sum_{k=1}^{n} f(a+k)=16\left(2^{n}-1ight)$\ $\Rightarrow f(a) \sum_{k=1}^{n} f(k)=16\left(2^{n}-1ight)$\ $\Rightarrow f(a)\left[2+2^{2}+\ldots \ldots . .+2^{n}ight]=16\left(2^{n}-1ight)$\ $\Rightarrow f(a) \cdot \frac{2\left(2^{n}-1ight)}{2-1}=16\left(2^{n}-1ight)$\ $\Rightarrow \quad f(a)=8=2^{3} \Rightarrow$ conclusion is $a=3$\