0

Is there any formular to this sum? $$ \binom{n}{1} + \binom{n+1}2 + \binom{n+2}3 + \cdots + \binom{n+m-1}m$$

This is a sum of left aligned pascal’s triangle.

For example, that is the sum of below numbers.(n=5, m=4)

[[ 1  1  1  1  1]
 [ 1  2  3  4  5]
 [ 1  3  6 10 15]
 [ 1  4 10 20 35]]

It looks like the sum is $$ \binom{m}1\binom{n}1 + \binom{m}2\binom{n}2 + \binom{m}3\binom{n}3 + \cdots + \binom{m}n\binom{n}n, \;\text{where } m>n$$ But I can't get a shorter form.

plhn
  • 631

2 Answers2

4

Add and subtract $\displaystyle \binom n0$ and use the formula $$\binom nr + \binom{n}{r+1}=\binom{n+1}{r+1}$$ To get

$$\left( \color{blue}{\binom n0} +\binom{n}{1} + \binom{n+1}2 + \binom{n+2}3 + \cdots + \binom{n+m-1}m \right) -\color{blue}{\binom n0} $$ $$= \binom{n+m}{m+1}-\color{blue}1$$

Jaideep Khare
  • 19,293
0

We have

$\binom {n-1}k = \binom{n}{k}-\binom{n-1}{k-1}$

So

$\binom {n}1 = \binom{n+1}{1}-\binom{n}{0} $

$\binom {n+1}2 = \binom{n+2}{2}-\binom{n+1}{1} $

...

$\binom {n+m-1}{m} = \binom{n+m}{m+1}-\binom{n+m-1}{m} $

Summarize above equations:

$\binom {n}1 + \binom {n+1}2 + ... + \binom {n+m-1}{m} = \binom{n+m}{m+1} -\binom{n+m-1}{m} + \binom{n+m-1}{m} - \binom{n+m-2}{m-1} + ... + \binom{n+1}{1}-\binom{n}{0}$

$=\binom{n+m}{m+1} - \binom{n}{0}$

Ronaldinho
  • 11