Let f : N → R be a function such that f(x+y)=2 f(x) f(y) for natural numbers x and y . If f(1)=2 ,

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1.Relation and Function

hard

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

$\sum \limits_{k=1}^{10} f(\alpha+k)=\frac{512}{3}\left(2^{20}-1\right)$ holds, is

$2$

$3$

$4$

$6$

(JEE MAIN-2022)

Solution

$f : N \rightarrow R , f ( x + y )=2 f ( x ) f ( y )$

$f (1)=2$,

$\sum_{ k =1}^{10} f (\alpha+ k )=2 f (\alpha) \sum_{ k =1}^{10} f ( k )$

$=2 f (\alpha)( f (1)+ f (2)+\ldots .+ f (10))$

From $(1)$

$f (2)=2 f ^{2}(1)=2^{3}$

$\left.f (3)=2 f (2) f (1)=2^{5}\right)$

$\vdots$

$f (10)=2^{9} f ^{10}(1)=2^{19}$

$f (\alpha)=2^{2 \alpha-1} ; \alpha \in N$

from $(2)$

$\sum_{ k =1}^{10} f (\alpha+ k )=2\left(2^{2 \alpha-1}\right)\left(2+2^{3}+2^{5}+\ldots .+2^{19}\right)$

$\frac{512}{3}\left(2^{20}-1\right)=2^{2 \alpha}\left(2 \frac{\left(2^{20}-1\right)}{3}\right)$

Hence $\alpha=4$

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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 $\sum \limits_{k=1}^{10} f(\alpha+k)=\frac{512}{3}\left(2^{20}-1\right)$ holds, is