No. of real roots.

[man111]

Let $f(x)$ be a polynomial of degree $n$, an odd positive integer, and has monotonic behaviour , then the number of real roots of the equation
$f(x)+f(2x)+......+f(nx) = \frac{1}{2} n(n+1)$ is equal to

[mihiv]

$f'(x)\geqslant 0$ (or $f'(x)\leqslant 0$)for all $x$,because f(x) is monotonous,let $P(x)=f(x)+f(2x)+\dots +f(nx)-\frac 12n(n+1)$,obviously $P'(x)\geqslant 0,x\in (-\infty,+\infty)$,then $P(x)$ is monotonous and has odd degree n and thus has only one real root.

[man111]

mihiv Thanks for nice explanation.
(I have also got 1 solution. using example.)