Finding number of roots of $1+2x+3x^2+...+(n+1)x^n=0$ Find the number of roots of the equation
$1+2x+3x^2+4x^3+....+(n+1)x^n=0$
where $n$ is even.
My attempt:
Let $f(x)=x+x^2+x^3+....+x^n$
Clearly,$x=0$ is a root and $f(x)=0$
cannot have a positive root.
$f(x)=x\left(1+x+x^2+...+x^n\right)$
How to determine whether the equation 
$1+x+x^2+x^3+...+x^n=0$ 
has real roots or not.
If I come to know that equation has certain number of roots then i can invoke Rolle's Theorem to find number of roots of given equation.
 A: We have $$ 1+2x+3x^2+...+(n+1)x^n = \sum_{k=0}^n (k+1)x^k = \sum_{k=0}^n \sum_{m=k}^n x^m = \sum_{k=0}^n \sum_{m=0}^{n-k} x^{m+k} =$$
$$= \sum_{k=0}^n x^k \sum_{m=0}^{n-k} x^m = \sum_{k=0}^n x^k \cdot \frac{x^{n-k+1}-1}{x-1} = \frac{1}{x-1} \sum_{k=0}^n {(x^{n+1}-x^k)} = $$
$$= \frac{(n+1)x^{n+1}}{x-1} - \frac{1}{x-1}\sum_{k=0}^n {x^k} = \frac{(n+1)x^{n+1}}{x-1} - \frac{1}{x-1}\cdot\frac{x^{n+1}-1}{x-1}=$$
$$=\frac{(n+1)x^{n+1}\cdot (x-1)-x^{n+1}+1}{(x-1)^2}=\frac{(n+1)x^{n+2}-(n+2)x^{n+1}+1}{(x-1)^2}$$
Thus we got the short formula for the sum which depends on n and x. Next since n is even then $n=2l$ and 
$$ \frac{(n+1)x^{n+2}-(n+2)x^{n+1}+1}{(x-1)^2} = \frac{(2l+1)x^{2l+2}-(2l+2)x^{2l+1}+1}{(x-1)^2} = $$
$$=\frac{(2l+1)x^{2l+2}-(2l+2)x^{2l+1}+1}{(x-1)^2} - (l+1)x^{2l}+(l+1)x^{2l}=$$
$$ = (l+1)x^{2l} + \frac{1}{(x-1)^2}\cdot((2l+1)x^{2l+2}-(2l+2)x^{2l+1}+1 -(l+1)x^{2l}(x-1)^2)= $$
$$ = (l+1)x^{2l}+\frac{lx^{2l+2}-(l+1)x^{2l}+1}{(x-1)^2}$$
Last fraction is very similar to the first formula. So we make some extra actions
$$(l+1)x^{2l}+\frac{lx^{2l+2}-(l+1)x^{2l}+1}{(x-1)^2} = (l+1)x^{2l}+\frac{l(x^2)^{l+1}-(l+1)(x^2)^l+1}{(x^2-1)^2}\frac{(x^2-1)^2}{(x-1)^2}=$$
$$ = (l+1)x^{2l}+(x+1)^2\sum_{k=0}^{l-1}(k+1)(x^2)^k$$
Since this sum consists of non-negative terms it is equal to zero only if each term is zero. Thus, given equation has no roots.
