Let f : N [] R be a function such that f(x + y) = 2f(x)f(y) for natural numbers x and y.

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

$$\sum\limits_{k = 1}^{10} {f(\alpha + k) = {{512} \over 3}({2^{20}} - 1)} $$

holds, is :

Accepted Answer

Given,

$f(x + y) = 2f(x)f(y)$

and $f(1) = 2$

For x = 1 and y = 1,

$f(1 + 1) = 2f(1)f(1)$

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

For x = 1, y = 2,

$f(1 + 2) = 2f(1)y(2)$

$$ \Rightarrow f(3) = 2\,.\,2\,.\,{2^3} = {2^5}$$

For x = 1, y = 3,

$f(1 + 3) = 2f(1)f(3)$

$$ \Rightarrow f(4) = 2\,.\,2\,.\,{2^5} = {2^7}$$

For x = 1, y = 4,

$f(1 + 4) = 2f(1)f(4)$

$$ \Rightarrow f(5) = 2\,.\,2\,.\,{2^7} = {2^9}$$ ..... (1)

Also given

$$\sum\limits_{k = 1}^{10} {f(\alpha  + k) = {{512} \over 3}({2^{20}} - 1)} $$

$$ \Rightarrow f(\alpha  + 1) + f(\alpha  + 2) + f(\alpha  + 3)\, + \,\,...\,\, + \,\,f(\alpha  + 10) = {{512} \over 3}({2^{20}} - 1)$$

$$ \Rightarrow f(\alpha  + 1) + f(\alpha  + 2) + f(\alpha  + 3)\, + \,\,....\,\, + f(\alpha  + 10) = {{{2^9}\left( {{{({2^2})}^{10}} - 1} \right)} \over {{2^2} - 1}}$$

This represent a G.P with first term = 29 and common ratio = 22

$$\therefore$$ First term $$ = f(\alpha  + 1) = {2^9}$$ ..... (2)

From equation (1), $f(5) = {2^9}$

$$\therefore$$ From (1) and (2), we get

$$f(\alpha  + 1) = {2^9} = f(5)$$

$$ \Rightarrow f(\alpha  + 1) = f(5)$$

$$ \Rightarrow f(\alpha  + 1) = f(4 + 1)$$

Comparing both sides we get,

$$\alpha  = 4$$

Solutions · 1