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### Diagonal Sum in Pascal's Triangle

```
Date: 04/02/2001 at 23:35:43
From: Gene Potter
Subject: Pascal's Triangle

Please explain how to find the sum of the reciprocals of the diagonals
of Pascal's triangle.
```

```
Date: 04/03/2001 at 19:54:43
From: Doctor Schwa
Subject: Re: Pascal's Triangle

Hi Gene,

so you mean, for instance,

1/1 + 1/3 + 1/6 + 1/10 + ...

or in more compact math language,

Sum [2/(n(n+1))]

The nifty trick here is to "un-common-denominatorify" this fraction.
That is, it now has a common denominator of n(n+1). What fractions
A/n + B/(n+1) would add up to make this answer? (This method is known
as partial fractions, and you can get help with it in the Dr. Math
archives by searching for "partial fractions.")

It turns out that 2/n - 2/(n+1) will do the trick.

That is, you can rewrite

1/1  = 2/1 - 2/2
1/3  = 2/2 - 2/3
1/6  = 2/3 - 2/4
1/10 = 2/4 - 2/5
:

Now the sum is easy to see: All the terms but 2/1 cancel, so the
infinite sum is 2/1. The finite sum of the first n terms will be:

2/1 - 2/(n+1)

What about the next diagonal? You could try a similar trick, but the
algebra is bound to get messy. Well, not really:

6/(n(n+1)(n+2)) = 3/n - 6/(n+1) + 3/(n+2)

If you have trouble doing that partial fractions trick, the archives
will help. Here is a good starting point:

What is a Partial Fraction?
http://mathforum.org/dr.math/problems/ishmael1.8.99.html

So again:

1/1  = 3/1 - 6/2 + 3/3
1/4  = 3/2 - 6/3 + 3/4
1/10 = 3/3 - 6/4 + 3/5
1/20 = 3/4 - 6/5 + 3/6
:

By combining the terms with like denominators, you can see that this
equals 3/1 - 3/2 in the long run; all the rest cancel, and the answer
is 3/2.

Now if you want a general formula for the k'th diagonal, I'd suggest
doing one or two more by hand this way (it's not that hard), and then
looking for a pattern and trying to prove it once you know what the
answer is already... or write us back. This is an interesting problem.
If you make some progress but get stuck somewhere along the line,

I'm sure the answer is known, and I could just look it up somewhere
along with a convenient trick for evaluating it, but it's more fun to
try to figure it out for ourselves.

Thanks for the fun problem,

- Doctor Schwa, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory
High School Sequences, Series

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