Roots of a summation

Question

I'm trying to solve an equation: $$\sum_{n=0}^b \left(\left(\frac{a+xn}{b}\right)\binom{b}{n}(-x)^{b-n}\right)=0$$ Where a and b are constants. I thought of solving it by using the binomial theorem. As you can see, the equation is very similar to the binomial expansion: $$\sum_{n=0}^b \left(\binom{b}{n}(-x)^{b-n}\right)=(1-x)^b$$ I was wondering if there was a way to somehow separate the $\left(\frac{a+xn}{b}\right)$ from the summation to solve the equation.

Answer 1

The two sums that you need are $f_b(x) =\sum_{n=0}^b \binom{n}{b}x^n $ and $g_b(x) =\sum_{n=0}^b n\binom{n}{b}x^n $. Your sum can be written in terms of these.

The first you should know from the binomial theorem.

For the second,

$\begin{array}\\ f_b'(x) &=\sum_{n=0}^b \binom{n}{b}(x^n)'\\ &=\sum_{n=0}^b \binom{n}{b}nx^{n-1}\\ &=\frac1{x}\sum_{n=0}^b \binom{n}{b}nx^{n}\\ &=\frac1{x}g_b(x)\\ \end{array} $

so $g_b(x) =xf_b'(x) $

and this gives you $g$.