Question

Let $\sum _{k=1}^{10}f(a+k)=16(2^{10}-1)$, where the function $f$ satisfies $f(x+y)=f\left(x\right)+f\left(y\right)$ for all natural numbers $x,y$ and $f\left(1\right)=2$. Then the natural number '$a$' is:

• A. $2$
• B. $3$
• C. $4$
• D. $16$

Answer 1

The correct option is B $3$

We have $f\left(1\right)=2\text{and}f(x+y)=f\left(x\right)+f\left(y\right)$.
$\Rightarrow f(1+1)=f\left(1\right)\cdot f\left(1\right)=2\cdot 2=2^{2}f(1+2)=f\left(1\right)\cdot f\left(2\right)=2\cdot 4=2^3⋮f(1+9)=2^{10}\Rightarrow f\left(x\right)=2^x\forall x\in N\Rightarrow 2^{a+1}+2^{a+2}+\cdots +2^{a+10}=16(2^{10}-1)\Rightarrow 2^a(2+2^2+\cdots +2^{10})=16(2^{10}-1)\Rightarrow 2^a\cdot \frac{2\cdot (2^{10}-1)}{1}=16(2^{10}-1)$
On comparing, we get
$a+1=4$
Hence, $a=3$.