Proving Pascal's Identity for more than two terms

Asked Jan 28 '23 at 16:25

Active Jan 28 '23 at 16:28

Viewed 24 times

This involves a slightly different version of Pascal's Identity but is still based on it.

Prove that $$ {a \choose k} + {a+1 \choose k} + \dots {a+n \choose k} = {a+n+1 \choose k+1} $$

I have tried to prove it using induction, but the algebra is quite cumbersome. Is there a more elegant solution?

edited Jan 28 '23 at 16:28

asked Jan 28 '23 at 16:25

TheMathPro

It's not true. For example, when $n=0$ or $n=1.$ – Thomas Andrews Jan 28 '23 at 16:34
In general, this sum is: $$\binom{a+n+1}{k+1}-\binom{a}{k+1}$$ You can prove that by induction. – Thomas Andrews Jan 28 '23 at 16:40
If you're looking for an elegant solution, I think combinatorial proofs are among the most elegant out there, and there are some in this site, look it up – acat3 Jan 28 '23 at 16:51

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