Question

Find $a$ for which $\sum _{k=1}^{n}f\left(a+k\right)=16\left(2^n-1\right)$, 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 $f\left(1\right)=2$.

• A. $2$
• B. $4$
• C. $5$
• D. $3$

Answer 1

Answer: (D)

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, $\sum _{k=1}^{n}f\left(a+k\right)$
$=f\left(a\right)\sum _{k=1}^{n}f\left(k\right)$
$=2^a\left(2^1+2^2+2^3+...+2^n\right)$
$=2^a\left\{\frac{2\cdot \left(2^n-1\right)}{2-1}\right\}$
$=2^{a+1}\left(2^n-1\right)...\left(i\right)$
But, we are given $\sum _{k=1}^{n}f\left(a+k\right)=16\left(2^n-1\right)$
$\Rightarrow 2^{a+1}\left(2^n-1\right)=16\left(2^n-1\right)$
$\Rightarrow 2^{a+1}=2^4$
$\Rightarrow a=3$