Question

$f$ is a function defined as $\sum _{k=1}^{n}f(a+k)=16(2^n-1)$ and $f(x+y)=f\left(x\right).f\left(y\right)$ and $f\left(1\right)=2$ then integral value of $a$

• A. $3$
• B. $0$
• C. $2$
• D. $1$

Answer 1

The correct option is A $3$

$f(x+y)=f\left(x\right)f\left(y\right)$ ....(1)
Put $x=1,y=0$ in (1)
$f\left(1\right)=f\left(1\right)f\left(0\right)$
$\Rightarrow f\left(0\right)=1=2^0$
Put $x=1,y=1$ in (1)
$f\left(2\right)=f\left(1\right)f\left(1\right)$
$\Rightarrow f\left(2\right)=2^2$
Put $x=2,y=1$ in (1)
$f\left(3\right)=f\left(2\right)f\left(1\right)$
$\Rightarrow f\left(3\right)=2^3$
Hence, $f\left(k\right)=2^k$ for all whole numbers.
Now, by (1), we can write
$f(a+k)=f\left(a\right)f\left(k\right)$
${\sum }_{k=1}^{n}f(a+k)=f\left(a\right){\sum }_{k=1}^{n}f\left(k\right)$
$\Rightarrow 16(2^n-1)=2^a(2+2^2+2^3+....+2^n)$
$\Rightarrow 16(2^n-1)=2^{a+1}(2^n-1)$
$\Rightarrow 16=2^{a+1}$
$\Rightarrow a=3$