# Simplify $\sum_{i=0}^{n-1} { {2n}\choose{i}}\cdot x^i$

I am trying to simplify an expression involving summation as follows:

$$\sum_{i=0}^{n-1} { {2n}\choose{i}}\cdot x^i$$

where $n$ is an integer, and $x$ is a positive real number.

At a first glance, I can see that

$$\sum_{i=0}^{2n} { {2n}\choose{i}}\cdot x^i = {(1+x)}^{2n}.$$

But what if in the case when $i$ goes from 0 to $n-1$?

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## 2 Answers

If you sum from $0$ to $n-1$, then no longer you get an easy espression. Instead, you can get the sum in terms of the hypergeometric function

$$\left( x+1 \right) ^{2\,n}-{2\,n\choose n}{x}^{n} {F (1,-n;\,n+1;\,-x)}\,.$$

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It's just a partial answer.

Let $S_1(x):=\sum_{i=0}^{n-1}\binom{2n}ix^i$, and $S_2(x):=\sum_{i=n+1}^{2n}\binom{2n}ix^i$. Then writing $j=2n-i$, we get $$S_2(x)=\sum_{j=0}^{n-1}\binom{2n}jx^{2n-j}=S_1(x^{-1})x^{2n}.$$ So $$(1+x)^{2n}=S_1(x)+\binom{2n}nx^n+S_2(x)=S_1(x)+\binom{2n}nx^n+S_1(x^{-1})x^{2n}.$$ Writing $f(x):=\frac{S_1(x)}{x^n}$, we get that $f$ satisfies the functional equation $$f(x)+f(x^{-1})=\left(1+\frac 1x\right)^n(1+x)^n-\binom{2n}n.$$

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