Cut the knot: learn to enjoy mathematics
A math books store at a unique math study site. Shopping at the store helps maintain the site. Thank you.
Learning Math Online
Sites for teachers
Sites for parents
Terms of use
Awards
Interactive Activities

CTK Exchange
CTK Wiki Math
CTK Insights - a blog
Math Help

III Millennium Olympiad

Games & Puzzles
What Is What
Arithmetic
Algebra
Geometry
Probability
Outline Mathematics
Make an Identity
Book Reviews
Stories for Young
Eye Opener
Analog Gadgets
Inventor's Paradox
Did you know?...
Proofs
Math as Language
Things Impossible
Visual Illusions
My Logo
Math Poll
Cut The Knot!
MSET99 Talk
Other Math sites
Front Page
Movie shortcuts
Personal info
Privacy Policy

Guest book
News sites

Recommend this site

Sites for parents

Education & Parenting

Manifesto  |  Bookstore  |  Contents  |  Amazon store  |  Term index  |  What changed?  |  Contact  |  Recommend
RSS Feed: Recent changes at CTK

Ways To Count

The purpose of this page is to present several proofs of an identity that involves binomial coefficients:

(1) C(n, 1) + 2·C(n, 2) + 3·C(n, 3) + ... + n·C(n, n) = n2n-1.

Four such proofs have been collected in a 1999 issue of Crux Mathematicorum by Jimmi Chui, then a secondary school student. I found two more elsewhere and present all six below.

Pure Algebra

Let S denotes the sum in (1). Using the basic property of the binomial coefficients C(n, k) = C(n, n - k) we have

 
S= 0·C(n, 0) + 1·C(n, 1) + 2·C(n, 2) + ... + n·C(n, n)
 = 0·C(n, n) + 1·C(n, n-1) + 2·C(n, n-2) + ... + n·C(n, 0)
 = n·C(n, 0) + (n-1)·C(n, 1) + (n-2)·C(n, 2) + ... + 0·C(n, n).

Adding the first and the last sums gives

 
2S= n·C(n, 0) + n·C(n, 1) + n·C(n, 2) + n·C(n, 3) + ... + n·C(n, n)
 = n·2n,

which immediately applies (1).

Mathematical Induction

We shall prove (1) by mathematical induction. For n = 1, there is nothing to prove. Assume (1) holds for some n = k:

(2) C(k, 1) + 2·C(k, 2) + 3·C(k, 3) + ... + k·C(k, k) = k2k-1.

If the left side (2) is denoted S(k), we have using the addition formula

 
S(k+1)= C(k+1, 1) + 2·C(k+1, 2) + 3·C(k+1, 3) + ... + (k+1)·C(k+1, k+1)
 = 1·(C(k, 0) + C(k, 1)) + 2·(C(k, 1) + C(k, 2)) + ... + k·(C(k, k-1) + C(k, k)) + (k+1)C(k, k)
 = 1·C(k, 0) + 3·C(k, 1) + 5·C(k, 2) + ... + (2k+1)C(k, k)
 = (C(k, 0) + C(k, 1) + C(k, 2) + ... + C(k, k)) + 2·S(k)
 = 2k + 2·k·2k-1
 = (k + 1)·2k.

Generating functions

Consider the generating function for the binomial coefficients:

  f(x) = C(n, 0) + C(n, 1)·x + C(n, 2)·x2 + ... + C(n, n)·xn

The derivative of which is

  f '(x) = C(n, 1) + 2·C(n, 2)·x1 + ... + n·C(n, n)·xn-1.

So that f '(1) = S(n). On the other hand, by the binomial theorem,

  f(x) = (1 + x)n,

with the derivative

  f '(x) = n(1 + x)n-1,

from which f '(1) = n2n-1.

For the record, what we found is that n(1 + x)n-1 is the generating function for the sequence {kC(n, k)}. This observation leads directly to the evaluation of the average in the binomial distribution.

Combinatorial Proof 1

Consider a set of n people. We shall answer the question, "In how many ways it is possible to select a (non-empty) committee and its chairperson from the set of n people?" We shall answer the question in two ways.

There are C(n, k) committees with k members (k ≥ 1) and, for each, there are k choices of a chairperson. Thus S(n) is the total number of such committees with a chair.

We may also start by selecting the chairperson first and then complementing the committee by regular members out of the remaining n-1 fellows. There are n choices for the chairperson and 2n-1 possibilities for other members, which gives the right hand side in (1).

Combinatorial Proof 2

The expression S(n)/2n [Benjamin & Quinn, p. 66] also has a combinatorial interpretation. It answers the question, "What is the average size of a subset of {1, 2, ..., n}?" Indeed, there are exactly C(n, k) subsets with k terms whereas 2n is the total number of the subsets.

On the other hand, pair each subset with its complement. The two together have n terms and the average of n/2. So that

  S(n)/2n = n/2,

which is equivalent to (1).

Formula for Binomial Coefficients

Recollect that [Zeitz, p. 215]

  C(n, k) =
n!
k!(n-k)!

Then

 
S(n)= ∑
k·n!
k!(n-k)!,

where the sum is taken from k = 1 to k = n. Thus

 
S(n)= ∑
k·n!
k!(n-k)!,
 = ∑
k·n!
k·(k-1)!(n-k)!,
 = ∑
n!
(k-1)!(n-k)!,
 = ∑
n·(n-1)!
(k-1)!(n - k)!
 = n·∑
(n-1)!
(k-1)!((n-1)-(k-1))!,

(The derivation might have been shortened by a reference to the well known formula kC(n, k) = nC(n-1, k-1), the absorbtion identity. The latter is verified directly from the definition of the binomial coefficients as was actually done above.)

Applying the binomial theorem.

 
S(n)= n·∑C(n-1, k-1)
 = n·(C(n-1, 0) + C(n-1, 1) + ... + C(n-1, n-1))
 = n·2n-1,

References

  1. A. T. Benjamin, J. J. Quinn, Proofs That Really Count: The Art of Combinatorial Proof, MAA, 2003
  2. P. Zeitz, The Art and Craft of Problem Solving, John Wiley & Sons, 1999

Pascal's Triangle and the Binomial Coefficients

Copyright © 1996-2010 Alexander Bogomolny

35707159Page copy protected against web site content infringement by Copyscape

Search:
Keywords:

Google
Web CTK