$$A_{r,s}=1+r+{r+1\choose2}+\cdots+{r+s-1\choose s}$$

How to find the general expression of $A_{r,s}$.

Somehow, I feel it is related to the coefficients of expansion of $(1+x)^r$, but I am not sure how they could be related.

  • $\begingroup$ This is a sum along a main diagonal of Pascal's Triangle. Neat trick: the sum along a diagonal going left to right is immediately below and left of the last summand. Hence, $A_{r,s}=$ ${r+s}\choose s$. $\endgroup$ Oct 18, 2022 at 1:20
  • $\begingroup$ This is called the Hockey Stick Identity. $\endgroup$
    – JBL
    Oct 18, 2022 at 14:45

1 Answer 1


Note that $A_{r,s}=\displaystyle \sum_{k=0}^{s} \binom{r+k-1}{k}$. Using Pascal's identity you can write the sum as $\displaystyle \sum_{k=0}^{s} \left(\binom{r+k}{k}-\binom{r+k-1}{k-1}\right)$ then apply a change of variables on the second summand $u=k-1$ and cancel out terms.


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