The unique root on (1,2) For any given $n\geq 2$, let $x^n=\sum\limits_{k=0}^{n-1}x^{k}$ be the equation, prove: there is only one real root which in (1,2). 

We have $x^n=\frac{1-x^n}{1-x}$, which rearranges to $0=\frac{1-x^n}{1-x}-x^n=\frac{1-x^n-x^n+x^{n+1}}{1-x}$.  For this to be zero (on $x>1$), $f(x)=x^{n+1}-2x^n+1$ must be zero.  We now have the following facts:


*

*$f(2)=2^{n+1}-2^{n+1}+1=1>0$.

*$f(1)=1-2+1=0$.

*$f'(x)=(n+1)x^{n}-2nx^{n-1}=0$ has two solutions, $x=0$ and $x=\frac{2n}{n+1}\in(1,2)$

*$f'(1)=(n+1)-2n=1-n<0$.
Combining (2) and (4) we know $f(x)<0$ immediately to the right of $1$.  By (1) we know $f(x)>0$ at $2$.  By IVT, $f(x)=0$ somewhere in $(1,2)$.  Hence $f(x)$ has two zeroes, at least, in $[1,2)$.  If there were a third one, then $f'(x)=0$ would have at least two solutions in $(1,2)$ by MVT.  But by (3) there is only one such solution, so there is at most one zero of $f(x)$ in $(1,2)$.
