<s>[INST] Number of real roots of $p_n(x)=1+2x+3x^2+....+(n+1)x^n$ if $n$ is an odd integer 
If $n$ be an odd integer. Then find the number of real roots of the polynomial equation $p_n(x)=1+2x+3x^2+....+(n+1)x^n$

$$
p_n(x)=1+2x+3x^2+....+(n+1)x^n\\
x.p_n(x)=x+2x^2+....nx^n+(n+1)x^{n+1}\\
p(x)[1-x]=1+x+x^2+....+x^n-(n+1)x^{n+1}\\
p(x)=\frac{1+x+x^2+....+x^{n}}{1-x}-\frac{(n+1)x^{n+1}}{1-x}=\frac{x^{n+1}-1}{x-1}\frac{1}{1-x}-\frac{(n+1)x^{n+1}}{1-x}\\
=\frac{x^{n+1}-1-(n+1)(x-1)x^{n+1}}{-(x-1)^2}=\frac{x^{n+1}-1+(n+1)x^{n+1}-(n+1)x^{n+2}}{-(x-1)^2}\\
=\frac{(n+2)x^{n+1}-(n+1)x^{n+2}-1}{-(x-1)^2}=\frac{1-(n+2)x^{n+1}+(n+1)x^{n+2}}{(x-1)^2}=0\\
\implies \boxed{(n+2)x^{n+1}-(n+1)x^{n+2}=1}
$$
I think I am stuck with my attempt, how can I find the real solutions ?
 [/INST] $$
p(x)=1+2x+3x^2+\ldots+(n+1)x^n;\\
xp(x)=x+2x^2+3x^3+\ldots+(n+1)x^{n+1};\\
(1-x)p(x)=1+x+x^2+\ldots+x^n-(n+1)x^{n+1}=\frac{1-x^{n+1}}{1-x}-(n+1)x^{n+1}\\
=\frac{1-x^{n+1}-(1-x)(n+1)x^{n+1}}{1-x}=0.\\
\Rightarrow1-x^{n+1}-(1-x)(n+1)x^{n+1}=0
$$
Note that $x=1$ is clearly not a solution of the original equation. As a result, the above equation is actually equivalant to your original equation, apart from having an extra root $x=1$.
Here is the equation we are trying to solve:
$$
1-x^{n+1}-(1-x)(n+1)x^{n+1}\equiv (n+1)x^{n+2}-(n+2)x^{n+1}+1=0\\
\Leftrightarrow(n+1)x^{n+2}=(n+2)x^{n+1}-1
$$
Since the original equation has no roots for $x\geq 0$, we now focus on the case $x<0$. When $x<0$, LHS is decreasing toward $-\infty$; RHS is increasing towards $\infty$. LHS>RHS for $x=0$. This tells us that the original equation has only one root.
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