TITLE: Combinatorial proofs of the following identities
QUESTION [7 upvotes]: I've been trying to find combinatorial proofs of the following two identities:
1: $\displaystyle\sum_{i=0}^{k} \binom{n}{i} = \sum_{i=0}^{k} \binom{n-1-i}{k-i} 2^i$ with $0 \le k \le n-1$
2: $\displaystyle\binom{2m}{2n} = \sum_{k=0}^{n} \binom{2n+1}{2k+1} \binom{m+k}{2n}$

For 1: The LHS is counting the number of subsets of size at most k from a set of size n.  The $2^i$ in the RHS makes me think of partitioning based on what elements can be considered from the full set then either including them or not, but I can't think of a way of doing this partition without overcounting and trying to interpret the binomial term hasn't helped.
For 2: Again, the LHS is simple enough but I'm lost on how to interpret the RHS.  Just from looking at it I feel like I should be considering some parity argument but haven't come up with anything else.

Any suggestions on how to proceed? Should I be looking for a more formal bijection?

REPLY [2 votes]: Here is another variation based upon the usage of the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write e.g.
\begin{align*}
  [z^k](1+z)^n=\binom{n}{k}
  \end{align*}

We start with (2) and obtain for $0\leq n\leq m$
  \begin{align*}
\sum_{k=2n-m}^m&\binom{2n+1}{2k+1}\binom{m+k}{2n}\tag{1}\\
&=\sum_{k=0}^{2m-2n}\binom{2n+1}{4n-2m+2k+1}\binom{2n+k}{2n}\tag{2}\\
&=\sum_{k=0}^{2m-2n}\binom{2n+1}{2m-2n-2k}\binom{-(2n+1)}{k}(-1)^k\tag{3}\\
&=\sum_{k=0}^{\infty}[z^{2m-2n-2k}](1+z)^{2n+1}[u^{k}](1+u)^{-(2n+1)}(-1)^k\tag{4}\\
&=[z^{2m-2n}](1+z)^{2n+1}\sum_{k=0}^\infty\left(-z^2\right)^k[u^k](1+u)^{-(2n+1)}\tag{5}\\
&=[z^{2m-2n}](1+z)^{2n+1}(1-z^2)^{-(2n+1)}\tag{6}\\
&=[z^{2m-2n}](1-z)^{-(2n+1)}\\
&=[z^{2m-2n}]\sum_{k=0}^\infty\binom{-(2n+1)}{k}(-z)^k\tag{7}\\
&=\binom{-(2n+1)}{2m-2n}\tag{8}\\
&=\binom{2m}{2n}\tag{9}
\end{align*}
  and the claim follows.

Comment:

In (1) we sum starting from $k=2n-m$, since $\binom{m+k}{2n}=0$ for $0\leq k<2n-m$.
In (2) we shift the index $k$ to start from zero.
In (3) we use the binomial identities $\binom{p}{q}=\binom{p}{p-q}$ and $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
In (4) we apply the coefficient of operator twice and extend to upper limit of the series to $\infty$ without changing anything since we are adding zeros only.
In (5) we use the linearity of the coefficient of operator and apply the rule $[z^{p+q}]A(z)=[z^p]z^{-q}A(z)$.
In (6) we apply the substitution rule with $u=-z^2$
\begin{align*}
  A(z)=\sum_{k=0}^\infty a_kz^k=\sum_{k=0}^\infty z^k [u^k]A(u)
  \end{align*}
In (7) we use the binomial series expansion
In (8) we select the coefficient of $z^{2m-2n}$.
In (9) we apply the identity $\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q$.


Using the same technique we show (1):
  \begin{align*}
  \sum_{i=0}^k\binom{n}{i}=\sum_{i=0}^k\binom{n-1-i}{k-i}2^i
  \end{align*}
We obtain
  \begin{align*}
\sum_{i=0}^k\binom{n-1-i}{k-i}2^i
&=\sum_{i=0}^\infty[z^{k-i}](1+z)^{n-1-i}2^i\\
&=[z^k](1+z)^{n-1}\sum_{i=0}^\infty\left(\frac{2z}{1+z}\right)^i\\
&=[z^k](1+z)^{n-1}\frac{1}{1-\frac{2z}{1+z}}\\
&=[z^k](1+z)^{n}\frac{1}{1-z}\\
&=[z^k]\sum_{i=0}^\infty z^i(1+z)^{n}\\
&=\sum_{i=0}^k [z^{k-i}](1+z)^{n}\\
&=\sum_{i=0}^k \binom{n}{k-i}\\
&=\sum_{i=0}^k \binom{n}{i}\\
\end{align*}
and the claim follows.