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I am studying Pascal's identity but I can't understand how the last step of the equation below is derived.

Note: The identity which is being used here is $\dbinom{n}{k}=\dbinom{n-1}{k}+\dbinom{n-1}{k-1}$ and in particular this is applied for the first terms in the left of every row.

\begin{array} \ \dbinom{n}{k} &=\dbinom{n-1}{k}+\dbinom{n-1}{k-1} \\ &=\dbinom{n-2}{k} +\dbinom{n-2}{k-1}+\dbinom{n-1}{k-1} \\ & = \dbinom{n-3}{k}+\dbinom{n-3}{k-1} +\dbinom{n-2}{k-1} +\dbinom{n-1}{k-1} \\ \end{array}

In the end we have :

$\dbinom{n}{k}=\dbinom{k-1}{k-1}+\dbinom{k}{k-1}+\dbinom{k+1}{k-1} +\cdots + \dbinom{n-1}{k-1}$

This last step really confuses me,I can't make sense of it.

By looking at the identity we have used for every beginning term , I would think that the first two binomal coefficients of the last step should have a form like $\dbinom{r-1}{k} +\dbinom{r-1}{k-1}$ ,but I don't have this form so the author must have used something else also...

Can someone help me understand why the last step look so ?

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You could write that first term as $\dbinom{k}{k}$, but note that $\dbinom{k}{k} = 1 = \dbinom{k-1}{k-1}$ so the adjustment is made as a nicety to make the lower expression more consistent across the formula.

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    $\begingroup$ Thanks for your response.(Couldn't the author point this at least or is this too trivial lol ?) $\endgroup$ – Mr. Y Jan 16 '16 at 20:58
  • $\begingroup$ @Mr.Y: I didn't find this initially obvious either, but some others might. $\endgroup$ – Daniel R. Collins Jan 17 '16 at 18:23

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