Number of real roots of $p_n(x)=1+2x+3x^2+....+(n+1)x^n$ if $n$ is an odd integer

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?

$$ p(x)=1+2x+3x^2+\ldots+(n+1)x^n;\\ xp(x)=x+2x^2+3x^3+\ldots+(n+1)x^{n+1};\\ (1-x)p(x)=1+x+x^2+\ldots+x^n-(n+1)x^{n+1}=\frac{1-x^{n+1}}{1-x}-(n+1)x^{n+1}\\ =\frac{1-x^{n+1}-(1-x)(n+1)x^{n+1}}{1-x}=0.\\ \Rightarrow1-x^{n+1}-(1-x)(n+1)x^{n+1}=0 $$ 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.