Question

Let $f:N\to R$ be a function such that $f(x+y)=2f\left(x\right)f\left(y\right)$ for natural numbers $x$ and $y$. If $f\left(1\right)=2$, then the value of $\alpha$ for which $\sum _{k=1}^{10}f(\alpha +k)=\frac{512}{3}(2^{20}-1)$ holds, is :

Answer 1

$f(x+y)=2f\left(x\right)f\left(y\right)$ & $f\left(1\right)=2$
$x=y=1$
$\begin{array}{c}\Rightarrow f\left(2\right)=2^3\\ x=2,y=1\\ \Rightarrow f\left(3\right)=2^5\end{array}\}f\left(x\right)=2^{(2x-1)}$
Now, $\sum _{k=1}^{10}f(\alpha +k)=\frac{512}{3}(2^{20}-1)$
$2\sum _{k=1}^{10}f\left(\alpha \right)f\left(k\right)=\frac{512}{3}(2^{20}-1)$
$2f\left(\alpha \right)\left[f\right(1)+f(2)+\cdots +f(10\left)\right]=\frac{512}{3}(2^{20}-1)$
$=2f\left(\alpha \right)[2+2^3+2^5+\cdots \text{upto}10\text{terms}]=\frac{512}{3}(2^{20}-1)$
$=2f\left(\alpha \right)\cdot 2\left(\frac{2^{20}-1}{4-1}\right)=\frac{512}{3}(2^{20}-1)$
$f\left(\alpha \right)=128=2^{2\alpha -1}$
$=2\alpha -1=7$
$\Rightarrow \alpha =4$