How does $\sum_{k=0}^{n-p} \sum_{i=0}^{k} C^{p-1+i}_i\cdot C^{n-(p+i)}_{k-i}\cdot(p+i)$ simplify to $p\cdot\sum_{k=0}^{n-p} C^{n+1}_k$?

Asked Jan 22 '22 at 10:10

Active Jan 22 '22 at 16:39

Viewed 77 times

Given equation: $$\sum_{k=0}^{n-p} \sum_{i=0}^{k} C^{p-1+i}_i\cdot C^{n-(p+i)}_{k-i}\cdot(p+i)$$ here

I can simplify this to $$\sum_{k=0}^{n-p} \sum_{i=0}^{k} p\cdot C^{p+i}_i\cdot C^{n-(p+i)}_{k-i}$$

But i can't go any further
I need to simplify this thing to:

$$p\cdot\sum_{k=0}^{n-p} C^{n+1}_k$$

How to reach here?

edited Jan 22 '22 at 16:39

RobPratt

asked Jan 22 '22 at 10:10

itti_da

Do you know generating functions? The inner sum is a convolution of two generating functions – Alan Abraham Jan 22 '22 at 16:50
1

See this question and there replace $l$ with $i$, $n$ with $n+1$, $j$ with $n+1-k$, $M$ with $p+1$, to get $\sum_{i=0}^{k} \binom{p+i}{i} \binom{n-p-i}{k-i} = \binom{n+1}{k}$. – Fabius Wiesner Jan 22 '22 at 17:52
What's $C_i^k$? – Diger Jan 22 '22 at 18:06
@diger that notation is equivalent to the binomial coefficient, $\binom{k}{i}$ – Alan Abraham Jan 22 '22 at 18:07
ok, thanks. Never seen it. – Diger Jan 22 '22 at 18:09
First, notice that $\binom{p+i-1}{i}(p+i)=p\binom{p+i}{i}$. – Alexander Burstein Jan 22 '22 at 18:42
@AlanAbraham I only know some very basics about generating functions. I don't know convolution – itti_da Jan 23 '22 at 09:48
@Aditi You can use the proofs in the post BillyJoe linked. The third answer is the generating function proof. – Alan Abraham Jan 23 '22 at 14:57

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