Post Number:#2 by Absur Khan Siam » Tue Feb 28, 2017 12:37 pm
$f^3(1) + f^3(2) + ... + f^3(n) = (\frac{f(n)f(n+1)}{2})^2...(i)$
$f^3(1) + f^3(2) + ... + f^3(n) + f^3(n+1)= (\frac{f(n+1)f(n+2)}{2})^2 ...(ii)$
$(ii) - (i) \Rightarrow f(n+1) = \frac{f^2(n+2)}{4} - \frac{f^2(n)}{4} = (\frac{f(n+2) + f(n)}{2})(\frac{f(n+2) - f(n)}{2})$
Since, $f(n+1) > 0 \Rightarrow f(n+2) > f(n)$
This , implies that $f(1) < f(2) < ... < f(n)$
Let, $f(n+2) = f(n) + k$
if $k < 2 , f(n+1) < f(n)$ ; contradiction
if $k > 2 , f(n) > f(n+2)$ ; contradiction
Thus , $k = 2 , f(n+1) = f(n) + 1$
$f(1) = 1$.By induction , we can say that $f(n) = n$.
$\therefore f(n) = n$.
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid