Let f : N rightarrow R be a function such tha | vedclass

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(C) Given $f(x+y) = 2f(x)f(y)$ and $f(1) = 2$.Let $g(x) = 2f(x)$. Then $g(x+y) = 2f(x+y) = 4f(x)f(y) = g(x)g(y)$.Since $g(1) = 2f(1) = 4 = 2^2$,we hav

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$4$

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(C) Given $f(x+y) = 2f(x)f(y)$ and $f(1) = 2$.Let $g(x) = 2f(x)$. Then $g(x+y) = 2f(x+y) = 4f(x)f(y) = g(x)g(y)$.Since $g(1) = 2f(1) = 4 = 2^2$,we have $g(x) = 2^{2x}$.Thus,$f(x) = \frac{1}{2} g(x) = \frac{1}{2} \cdot 2^{2x} = 2^{2x-1}$.Now,$\sum_{k=1}^{10} f(\alpha+k) = \sum_{k=1}^{10} 2^{2(\alpha+k)-1} = 2^{2\alpha-1} \sum_{k=1}^{10} 2^{2k} = 2^{2\alpha-1} \cdot 4 \cdot \frac{4^{10}-1}{4-1} = 2^{2\alpha+1} \cdot \frac{2^{20}-1}{3}$.We are given $\sum_{k=1}^{10} f(\alpha+k) = \frac{512}{3}(2^{20}-1) = \frac{2^9}{3}(2^{20}-1)$.Equating the two expressions: $2^{2\alpha+1} \cdot \frac{2^{20}-1}{3} = \frac{2^9}{3}(2^{20}-1)$.$2^{2\alpha+1} = 2^9 \implies 2\alpha+1 = 9 \implies 2\alpha = 8 \implies \alpha = 4$.

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

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