Question

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

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

Answer: (C)

$f:N\to R,f(x+y)=2f(x)f(y).....$(1)
$f(1)=2$,
$\sum _{k=1}^{10}f(\alpha +k)=2f(\alpha )\sum _{k=1}^{10}f(k)$
$=2f(\alpha )(f(1)+f(2)+\dots ..+f(10))\dots ..(2)$
From $(1)$
$f(2)=2f^2(1)=2^3$
$f(3)=2f(2)f(1)=2^5$
$⋮⋮$
$f(10)=2^9{f}^{10}(1)=2^{19}$
$f(\alpha )=2^{2\alpha -1};\alpha \in N$
from $(2)$
$\sum _{k=1}^{10}f(\alpha +k)=2\left(2^{2\alpha -1}\right)\left(2+2^3+2^5+\dots .+2^{19}\right)$
$\frac{512}{3}\left(2^{20}-1\right)=2^{2\alpha }\left(2\frac{\left(2^{20}-1\right)}{3}\right)$
Hence $\alpha =4$