<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] You have found that $p(x)$ is a rational function. It is zero if and only if its numerator is zero. Its numerator is a polynomial of the form $Ax^{n+2}+Bx^{n+1}+C$, where $A,B,C$ are constants. The derivative of that numerator is of the form $Dx^{n+1}+Ex^n=x^n(Dx+E)$ for some constants $D,E$. It's easy to see how many real roots that derivative has. Then Rolle's Theorem tells you something about how many real roots the original numerator has. Can you fill in the details and draw a conclusion?
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