Question

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

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

Answer: (C)

$f : N \rightarrow R , f ( x + y )=2 f ( x ) f ( y ) .....$(1)
$f(1)=2$,
$\displaystyle\sum_{ k =1}^{10} f (\alpha+ k )=2 f (\alpha) \displaystyle\sum_{ k =1}^{10} f ( k )$
$=2 f (\alpha)( f (1)+ f (2)+\ldots . .+ f (10)) \ldots . .(2)$
From $(1)$
$f (2)=2 f ^{2}(1)=2^{3}$
$f (3)=2 f (2) f (1)=2^{5}$
$\vdots \,\,\,\,\, \vdots$
$f (10)=2^{9} f ^{10}(1)=2^{19}$
$f (\alpha)=2^{2 \alpha-1} ; \alpha \in N$
from $(2)$
$\displaystyle\sum_{ k =1}^{10} f (\alpha+ k )=2\left(2^{2 \alpha-1}\right)\left(2+2^{3}+2^{5}+\ldots .+2^{19}\right)$
$\frac{512}{3}\left(2^{20}-1\right)=2^{2 \alpha}\left(2 \frac{\left(2^{20}-1\right)}{3}\right)$
Hence $\alpha=4$