Finite sum of binomial coefficients

Question

Is there a way to calculate:$\sum_{i=0}^{k} {2k+1\choose k-i}$ using only:

• symmetry;
• pascal's triangle;
• one of these sums: $$\sum_{i=0}^{k} {n+i\choose i}={n+k+1\choose k}$$ and $${p\choose p}+{p+1\choose p}+\dots+{n\choose p}={n+1\choose p+1}.$$ I am not sure, because I do not see a way to get same lower number in all collectors so I can use the second sum. Also I tried using symmetry but not helpful really.

Answer 1

$$S=\sum_{i=0}^{k} {2k+1 \choose k-i}=\sum_{j=k}^{0} {2k+1 \choose j}=\sum_{l=0}^{k} {2k+1 \choose l} \tag1$$ We know that $$2^{2k+1}=\sum_{p=0}^{2k+1} {2k+1 \choose p} =\sum_{p=0}^{k} {2k+1 \choose p}+ \sum_{p=k+1}^{2k+1} {2k+1 \choose p}=S+S'\tag2$$ In the second sum let $q=p-k-1$, then $$S'=\sum_{q=0}^{k} {2k+1 \choose q+k+1}=\sum_{q=0}^{k} {2k+1 \choose k-q} \tag3.$$ Let $j=k-q$, then $$S'=\sum_{j=0}^{j=k} {2k+1 \choose j}=S \tag4$$ From (2), we get $$S+S'=2S=2^{2k+1} \implies S=2^{2k}=4^k$$

Answer 2

From symmetry you have got

$$\sum_{i=0}^k \binom{2k+1}{k-i} = \sum_{i=1}^{k+1}\binom{2k+1}{k+i}$$

Summing them up we have

$$\begin{align}&\quad\sum_{i=0}^k \binom{2k+1}{k-i} + \sum_{i=1}^{k+1}\binom{2k+1}{k+i}\\&=\binom{2k+1}{k}+\binom{2k+1}{k-1}+\cdots + \binom{2k+1}{0}+\binom{2k+1}{k+1}+\binom{2k+1}{k+2}+\cdots+\binom{2k+1}{2k+1}\\&=\sum_{i=0}^{2k+1}\binom{2k+1}{i}\end{align}$$

The last sum is a famous sum involving the binomial theorem.