Let n and k be positive integers such that \(n<10^6\) and \(\binom{13}{13} + \binom{14}{13} + \binom{15}{13} + \dots + \binom{52}{13} + \binom{53}{13} + \binom{54}{13} = \binom{n}{k}\)

What is the ordered pair (n, k)?

mathmathj28 Feb 25, 2020

#1**+1 **

\(\text{Using the hockey stick identity}\\ \sum \limits_{n=13}^{54}\dbinom{n}{13} = \dbinom{54+1}{13+1} = \dbinom{55}{14}\)

Rom Feb 25, 2020

#2**+1 **

oh that makes sense to use the hockey stick identity. Thanks for the help!!

mathmathj28
Feb 25, 2020