Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $n-1$?

share|improve this question
add comment

2 Answers

up vote 1 down vote accepted

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

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

share|improve this answer
add comment

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 $$f(x)+f(x^{-1})=\left(1+\frac 1x\right)^n(1+x)^n-\binom{2n}n.$$

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.