I want to prove
$$\sum _{r=0}^{n}\binom{n+r}{r}=\binom{2n+1}{n}$$
I have tried expanding the LHS and adding and substracting in order to force the Pascal's Triangle identity $\binom{\:n+1}{r+1\:}=\binom{\:n}{\:r}+\binom{n}{r+1}$. But this led nowhere.
I also considered multiplying both sides by $x^r$, thus making the LHS $(1+x)^{n-r}$. The RHS fails.
What else can I do? Am how am I supposed to know/think?
Use $${n+1 \choose k}={n \choose k-1}+{n \choose k}~~~~(1)$$ successively, $$S=\sum_{r=0}^{n} {n+r \choose r}= \sum_{r=0}^{n} {n+r \choose n}=$$ $$S={n \choose n}+{n+1 \choose n}+{n+2 \choose n}+{n+3 \choose n}+....+{2n \choose n}$$ Add ${n \choose n+1}=0$ as the first term in above and use (1) to get $$S={n+1 \choose n+1}+{n+1 \choose n}+{n+2 \choose n}+{n+3 \choose n}+....+{2n \choose n}.$$ Combining the first two terms and using (1) again, we get $$S={n+2 \choose n+1}+{n+2 \choose n}+{n+3 \choose n}+....+{2n \choose n}.$$ Keep on reducing first two terms using (1) successively to get $$S={2n+1 \choose n+1}={2n+1 \choose n}$$
Minor variation of hockey stick identity.
First, see the diagram near the start of the cited article.
For your conjecture, the hockey stick is facing the other direction. The idea is that each number in Pascal's Triangle is the sum of the two numbers above it.
Easiest illustration, that will convince you that it true for any $n \in \Bbb{Z^+}$ is to consider $n = 4.$
Draw Pascal's Triangle through the 9th row.
To prove: $$\binom{4}{0} + \binom{5}{1} + \binom{6}{2} + \binom{7}{3} + \binom{8}{4} = \binom{9}{4}.$$
Note that :
The same approach will work for any $n$. That is:
