Simplifying $\sum_{k=0}^b{n+k \choose n+k-a}$

Question

Let $a,b,n$ positive integers such that $a \leq n$. Can we simplify this formula?

$\sum_{k=0}^b{n+k \choose n+k-a}={ n \choose n-a } + { n+1 \choose n+1-a } + \dots + {n+b \choose n+b-a}$

I was stuck at the beginning except for the thought to use the hockey-stick identity somehow like mentioned in the comments.

see https://math.stackexchange.com/questions/870219/sum-of-k-combination-with-repetitions — Albus Dumbledore, Dec 21 '20 at 11:04
You could also have a look at Proof of the Hockey-Stick Identity: $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$ and other questions linked there. — Martin Sleziak, Dec 25 '20 at 07:39

score 3 · Accepted Answer · answered Dec 21 '20 at 10:22

3

$$S=\sum_{k=0}^{b} {n+k \choose n+k-a}=\sum_{k=0}^{b} {n+k \choose a}=[x^a]\sum_{k=0}^{b} (1+x)^{n+k}$$ $$S=[x^a] (1+x)^n \frac{(1+x)^{b+1}-1}{1+x-1}=[x^{a+1}]~((1+x)^{n+b+1}-(1+x)^n).$$ $$S={n+b+1 \choose a+1}-{n \choose 1+a}.$$ Here $[x^k]g(x)$ means the coefficient of $x^k$ in $g(x).$

answered Dec 21 '20 at 10:22

Z Ahmed

43,235

shouldn't it be $ x^{a-1}$? why did it become a+1? ( Line-2) – tryst with freedom Dec 21 '20 at 10:25
It is because of $x=1+x-1$ down. Check that $x^k=[x^{k+1}]g(x)$ – Z Ahmed Dec 21 '20 at 10:27

score 1 · Answer 2 · answered Dec 21 '20 at 10:43

Hint : Make use of the simple Pascal's identity, i.e. : $~\dbinom{n}{k} + \dbinom{n}{k+1} = \dbinom{n+1}{k+1}$

\begin{align*} \sum_{k=0}^b \dbinom{n+k}{n+k-a} &= \sum_{k=0}^b \dbinom{n+k}{a}\\ &= \dbinom{n}{a+1} + \dbinom{n}{a} + \dbinom{n+1}{a} + \cdots + \dbinom{n+b}{a} - \dbinom{n}{a+1}\\ &= \dbinom{n+1}{a+1} + \dbinom{n+1}{a} + \dbinom{n+2}{a} + \cdots + \dbinom{n+b}{a} - \dbinom{n}{a+1}\\ &\hspace{1.4cm}\vdots\hspace{2.2cm}\vdots\\ &= \dbinom{n+b}{a+1} + \dbinom{n+b}{a} - \dbinom{n}{a+1}\\ &= \dbinom{n+b+1}{a+1} - \dbinom{n}{a+1} \end{align*}

Hope it helps.

Simplifying $\sum_{k=0}^b{n+k \choose n+k-a}$

2 Answers2