I discovered that most elegant proof of a binomial identity is a combinatorial proof in many cases. I worked out reduction formula 's general version : $$C(m,m) + C(m+1,m) + ... C(n,m) = C(n+1,m+1)$$ I proved this by considering A set of $n+1$ elements and classifying it as it contains the elements of a set containing $n-m$ elements and exhausting all the possibilities. Then stands another problem which is $$C(n,m) +2C(n-1,m) ... + (n-m+1)C(m,m) = C(n+2,m+2)$$ This I proved by arranging its terms as triangular array of coefficients and repeatedly applying the general reduction formula (as I call it). Now stands the question of direct tricky combinatorial proof of this. I tried some manipulation and terms turn out to be similar to case Vandermonde identity except that the no. of elements in two sets are varying but sum of the two is same. Any elegant proofs are welcome. Also let me know what is standard names of the identities I mentioned. $C(n,r)$ denotes no. of combinations of $n$ elements taken $r$ at a time.
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The first identity is sometimes referred to as the hockey-stick identity. The second identity is as you correctly noticed a repeated application of the first. See also. – JMoravitz Sep 12 '17 at 18:05
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Did you mean to write $(n - m + 1)C(m, m)$? – N. F. Taussig Sep 12 '17 at 18:30
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Yes I meant that edited the question. – vishal mishra Sep 13 '17 at 03:45