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Answer by Felix Marin for Combinatorial Identity with powers of 2
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View ArticleAnswer by Markus Scheuer for Combinatorial Identity with powers of 2
Here is a variation which transforms the left-hand side into the right-hand side. In order to do so we use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way...
View ArticleAnswer by Max for Combinatorial Identity with powers of 2
Okay so it's not actually hard to get either $f(X)$ or $g(X)$, you just need to keep on using the well known generating function which I mentioned earlier. Above, we have already shown that the...
View ArticleAnswer by Angina Seng for Combinatorial Identity with powers of 2
Your $f(X)$ is$$\sum_{i=m}^\infty{i\choose m}\frac{X^{2i+1}}{(1-X)^{2i+2}}=\frac{X}{(1-X)^2}\frac{(X^2/(1-X)^2)^m}{(1-X^2/(1-X)^2)^{m+1}}$$(using the principle in your previous line). This simplifies...
View ArticleCombinatorial Identity with powers of 2
The following problem is from MOP 2007.Prove that$$\sum_{i=0}^n\binom{n}{2i+1}\binom{i}{m}=2^{n-2m-1}\binom{n-m-1}{m}$$I tried to approach this using generating functions:Let $f(X),g(X)$ be defined...
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