Expert: Paul Klarreich - 3/20/2009

Question
5. The sum:

1/(1!2009!) + 1/(3!2007!) + 1/(5!2005!) + ... + 1/(2005!5!) + 1/(2007!3!) + 1/(2009!1!)

can be written in the form (2^a)/(b!), where a and b are positive integers. Find a + b.

Answer
Questioner:   maegan
Country:  Philippines
Category:  Calculus
Private:  No
 
Subject:  expansion and series of factorial
Question:  5. The sum:

1/(1!2009!) + 1/(3!2007!) + 1/(5!2005!) + ... + 1/(2005!5!) + 1/(2007!3!) + 1/(2009!1!)

can be written in the form (2^a)/(b!), where a and b are positive integers. Find a + b.
....................................
Sorry -- I don't have a proof for you.  But I note the following:

The coefficients for (x + y)^2 are  1  2   1
The coefficients for (x + y)^4 are  1  4  6  4   1
The coefficients for (x + y)^6 are  1  6 15  20  15 6   1

In each case, if you partition it into the 'odd' and 'even' coefficients, the sums are equal.  So, for example:

(1 + 1)^6 =   1  + 6  + 15  + 20  + 15 + 6 + 1

and the 'even' sum is 2^6/2 = 2^5.

But to get your sum, you divide by  6! because the binomial coefficients, such as the 15, are of the form:
6!
----
2!4!

and your sum only has the bottom.

So if you divide by 6!, you have this sum:
 6!
-----
0!6!
PLUS
 6!
-----
1!5!
PLUS
 6!
-----
2!4!
PLUS
 6!
-----
3!3!
PLUS
 6!
-----
4!2!
PLUS
 6!
-----
5!1!
PLUS
 6!
-----
6!0!
--------------
Which is just like your sum, after deleting the ones with even numbers on the bottom.

And that would be  2^(6-1)/6!.

So your sum looks like  2^(2010 - 1)/2010!

= 2^(2009)/2010!

Now you'll have your a and b.

Maybe some time I'll have the proof for you.

========================================
A number of things intervened, but here is the proof:

Your sum:
  1          1          1              1          1          1      
-------- + -------- + ------- + ... + -------- + -------- + -------
1!2009!    3!2007!    5!2005!         5!2005!    3!2007!    1!2009!

Multiply the tops of all of those by  2010!  and you have:

C(2010,1) + C(2010,3) + ... + C(2010,2007) + C(2010,2009)

Now C(n,r) = C(n-1,r-1) + C(n-1,r) is a well-known property of the binomial coefficients, and we can apply that:

C(2010,1)    = C(2009,0)    + C(2009,1)
C(2010,3)    = C(2009,2)    + C(2009,3)
C(2010,5)    = C(2009,4)    + C(2009,5)
...
C(2010,2005) = C(2009,2004) + C(2009,2005)
C(2010,2007) = C(2009,2006) + C(2009,2007)
C(2010,2009) = C(2009,2008) + C(2009,2009)
---------------------------------------------- EQUALS:

C(2009,0) + ... + C(2009,2009)

= 2^2009

So the original is  2^2009/2010!