Let f: N R be a function satisfying the functional equation:f(x + y) = 2f(x)f(y)for all natural numbers x and y.If f(1) = 2, find the value of such that:_k = 1^10 f( + k) = 5123( 2^20 - 1 ) holds is

p q^2 x = q r y = p^2 r z Now 3 x y-3=12 \\ x y=76=q rp q^2=rp q \\ 7 p q=6 r 7(r-1) =6 r \\ r=7F and \\&nbs

Let f:N arrow 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}\mspace{2mu} f(\alpha + k) = \frac{512}{3}( 2^{20} - 1 ) holds, is holds, is: Sequence & Series (Mathematics)

$p q^2 \log x = q r \log y = p^2 r \log z$

Now $\frac{3 \log x}{\log y}-3=\frac{1}{2} \\ \Rightarrow \frac{\log x}{\log y}=\frac{7}{6}=\frac{q r}{p q^2}=\frac{r}{p q} \\ \Rightarrow 7 p q=6 r \Rightarrow 7(r-1) =6 r \\ \Rightarrow r=7F$ and $\\ \frac{3 \log y}{\log z}-3=1 \\ \Rightarrow \frac{\log y}{\log z}=\frac{4}{3}=\frac{p^2 r}{q r}=\frac{p^2}{q} \Rightarrow 3 p^2=4 q \\$ Also $\\ \frac{7 \log z}{\log x}-3=\frac{3}{2} \\ \Rightarrow \frac{7 \log z}{\log x}=\frac{9}{2}=\frac{7 p q^2}{p^2 r}=\frac{7 q^2}{p r} \\ \Rightarrow \frac{7 q^2}{p \times 7}=\frac{9}{2} \\ \Rightarrow \frac{q^2}{p}=\frac{9}{2} \Rightarrow 2 q^2=9 p \Rightarrow 4 q^4=81 p^2 \\ \Rightarrow 4 q^4=27 \times 3 p^2=27 \times 4 q \\ \Rightarrow q^3=27 \Rightarrow q=3 \\$ We have $r=p q+1 \Rightarrow 7=3 p+1 \Rightarrow p=2 \\ r-p-q=7-2-3=2$

Explanation

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