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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:

Given that
$f\left(x+y\right) = f\left(x\right)\cdot f\left(y\right)$ and $f\left(1\right) = 2$
Therefore, $f\left(2\right) $
$= f\left(1+1\right)= f\left(1\right)\cdot f\left(1\right) $
$= 2^{2}$
$ f\left(3\right) = f\left(1+2\right) $
$= f\left(1\right)\cdot f\left(2\right)$
$ = 2^{3}$ and so on.
continuing the process, we obtain
$f\left(k\right) = 2^{k}$ and $f\left(a\right) - 2^{a} $
Now, $\sum\limits_{k=1}^{n}f\left(a+k\right) $
$= f\left(a\right) \sum\limits _{k=1}^{n} f\left(k\right) $
$= 2^{a}\left(2^{1}+2^{2}+2^{3} + ...+ 2^{n}\right) $
$ = 2^{a}\left\{\frac{2\cdot\left(2^{n}-1\right)}{2-1}\right\}$
$ = 2^{a+1}\left(2^{n } -1\right)\quad...\left(i\right)$
But, we are given $\sum\limits_{k=1}^{n} f\left(a+k\right) = 16\left(2^{n}-1\right)$
$ \Rightarrow 2^{a+1}\left(2^{n}-1\right) = 16\left(2^{n}-1\right)$
$\Rightarrow 2^{a+1} = 2^{4}$
$\Rightarrow a=3$