I have a question regarding the zeros of the following polynomial, based on partial rows of pascal's triangle,

$$f(x)=\sum_{k=a}^n\binom{n}{k}(-1)^{2k+1}x^k,$$

where $a,n\in \mathbb{Z}^+,n>a.$

Specifically, I'm wondering if there exists zeros for $f$ when $x \in (0,1)$. Obviously, when $a=0$, the zeros occur at $x=1$, but as we increase $a$, I haven't found any solutions for $x\in (0,1)$.

For any $a \geq 0,$ does there exist $x \in (0,1)$ such that $f(x)=0$?

Potential leads include the Nörlund–Rice integrals.

http://en.wikipedia.org/wiki/N%C3%B6rlund%E2%80%93Rice_integral

I have a question regarding the zeros of the following polynomial, based on partial rows of pascal's triangle,

$$f(x)=\sum_{k=a}^n\binom{n}{k}(-1)^{k}x^k,$$

where $a,n\in \mathbb{Z}^+,n>a.$

Specifically, I'm wondering if there exists zeros for $f$ when $x \in (0,1)$. Obviously, when $a=0$, the zeros occur at $x=1$, but as we increase $a$, I haven't found any solutions for $x\in (0,1)$.

For any $a \geq 0,$ does there exist $x \in (0,1)$ such that $f(x)=0$?

Potential leads include the Nörlund–Rice integrals.

http://en.wikipedia.org/wiki/N%C3%B6rlund%E2%80%93Rice_integral