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26

MULTIPLICATION OF SUMS

A proof of the binomial theorem

THE BINOMIAL THEOREM gives the coefficients in the product of n equal binomials:

(x + a)n = (x + a)(x + a)· · · (x + a).

If we actually multiplied the 4 factors of

(x + a)4,

then before adding the like terms, we would find terms in

x4, x3a, x2a2, xa3, and a4.

The binomial theorem tells how many terms there are of each kind. Those binomial coefficients, the theorem states, are the combinatorial numbers.  To prove that, we will first consider the multiplication of any sums; for example,

(x + 1)(x + 2)(x + 3).

The final product will be a polynomial of the 3rd degree:

ax3 + bx2 + cx + d.

As with the binomial theorem, the question is:  What are the coefficients?

They will be some function of the constants 1, 2, 3.  For the moment, we will simply show the result:

(x + 1)(x + 2)(x + 3)  =  x3 + (1 + 2 + 3)x2 + (1· 2 + 1· 3 + 2· 3)x + 1· 2· 3
 
   =  x3 + 6x2 + 11x + 6.

The coefficient of x3 is simply 1.  The coefficient of x2 is the sum of the combinations of  1, 2, 3, taken one at a time.

The coefficient of x is the sum of the combinations of  1, 2, 3, taken two at a time.

And the constant term is their combination taken three at a time.

Why those combinations?  We will see why as we continue.  For the moment, the student should attempt this problem.

Problem 1.   Multiply out  (x + 3)(x + 4)(x − 1)  by taking the correct combinations of the integers.

To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").

  x3 + (3 + 4 − 1)x2 + (3· 4 + 3· −1 + 4· −1)x + 3· 4· −1
 
 =  x3 + 6x2 + 5x − 12

Let us now begin again, and analyze the multiplication of these elementary sums:

(x + y)(a + b + c).

Upon multiplying, we would find six terms.  Each term will contain two factors, namely one letter from each factor:

xa + xb + xc + ya + yb + yc.

Therefore, we can write the product of the following --

(x + y)(a + b)(m + n)

-- simply by writing the sum of all combinations of one letter from each factor.

xam + xan + xbm + xbn + yam + yan + ybm + ybn.

Each term in the product consists of three factors:  one from each binomial.

Note that there are a total of 23 or 8 terms.  In general:

Multiplication of n binomials produces 2n terms.

For, multiplication of two binomials gives 4 terms:

(p + q)(m + n) =  pm + pn + qm + qn.

If we multiply those with a binomial, we will have 8 terms; those multiplied with a binomial will produce 16 terms; and so on.

Now consider these four binomial factors, in which each of the constants, a, b, c, d, is different::

(x + a)(x + b)(x + c)(x + d).

And let us compare it with

(x + a)4 = (x + a) (x + a) (x + a) (x + a),

in which each constant is the same.

In this product --

(x + a)(x + b)(x + c)(x + d)

-- each of the 24 terms will consist of four factors: one from each binomial; hence we expect to find terms such as

xbcd, axcx, xxxd,

and so on.  That is, we will find terms in x,  x2, x3, and x4.  It remains to determine how many of each.

Now, how is a term x4 produced?  It will be produced by taking x from each factor.  xxxx = x4.  There is only one such term.  The coefficient of x4 is 1.

Similarly, then, in the expansion of

(x + a)4

the coefficient of x4 will be 1.

(x + a)(x + b)(x + c)(x + d)

Next, terms with x3 are formed by taking  x  from any three factors, in every possible way, and the letter from the remaining factor.

axxx + xbxx + xxcx + xxxd = (a + b + c + d)x3.

The coefficient of x3, therefore, is the sum of the combinations of  a, b, c, d  taken 1 at a time:  a + b + c + d.  How many such combinations are there?

4C1: The number of combinations of 4 things taken 1 at a time.

In the expansion of

(x + a) (x + a) (x + a) (x + a)

then, terms with x3a will be produced by taking x from three factors in every possible way, and a from the remaining factor. Again, there will be 4C1 such terms.

axxx + xaxx + xxax + xxxa = 4x3a.

The coefficient of x3a will be 4C1 itself: 4.

(x + a)(x + b)(x + c)(x + d)

Next, terms with x2 will come from taking x from two factors, in every possible way, and the letters from the remaining two. How many such terms will there be?  4C2: The number of ways of choosing 2 things -- 2 letters-- from 4.  The sum of those like terms will be

(ab + ac + ad + bc + bd + cd)x2.

As for terms in x2a2 in the expansion of

(x + a) (x + a) (x + a) (x + a),

the only difference is that all the constants are the same. There will be 4C2 such terms:

(aa + aa + aa + aa + aa + aa)x2.

The coefficient of x2a2 will be 4C2: 6.

(x + a)(x + b)(x + c)(x + d)

A term in x will be produced by taking x from 1 factor and the letter from the remaining 3:

(abc + abd + acd + bcd)x.

There will be 4C3 or 4 ways of doing that; of choosing 3 letters from 4.

In the expansion of

(x + a) (x + a) (x + a) (x + a),

there will similarly be 4C3 terms in a3x:

(aaa + aaa + aaa + aaa)x.

The coefficient of a3x will be 4C3.

(x + a)(x + b)(x + c)(x + d)

Finally, the constant term will be produced by taking the letter from each of the 4 factors. There is 4C4 -- 1 -- way of doing that. The constant term will be

abcd.

As for the constant term in the expansion of

(x + a) (x + a) (x + a) (x + a).

again there will be 4C4 -- 1 -- such term. The constant term will be

aaaa = a4.

We have found:

(x + a)(x + b)(x + c)(x + d)

 =  x4  + (a + b + c + c)x3 + (ab + ac + ad + bc + bd + cd)x2
 
  + (abc + abc + acd + bcd)x + abcd.
(x + a)4 = 4C0x4 + 4C1ax3 + 4C2a2x2 + 4C3a3x + 4C4a4
 
  = x4 + 4ax3 + 6a2x2 + 4a3x + a4.

We have in this way accounted for the binomial coefficients in the epansion of (x + a)n, with n = 4.  They are 1  4  6  4  1.

The binomial coefficients are how many terms there are of each kind.

The result is general.  The binomial theorem states that in the expansion of (a + b)n, the coefficients are the combinatorial numbers nCk , where k -- the exponent of b -- successively takes the values 0, 1, 2, . . . , n.

  Each term in theexpansion will have this form:          nExclamation!       
(nk)Exclamation! kExclamation!
 ankbk .

Problem 2.   Imagine multiplying out  (x + y + z)(a + a + c).

a)  How many terms would there be?   32 = 9

b)  Each term would consist of how many factors?   Two

Problem 3.   Imagine multiplying out  (x + a)(x + a)(x + c)(x + d).

a)   How many terms would there be?   24 = 16

Thus a product of n binomials consists of how many terms?   2n

b)  Each term would consist of how many factors?   Four

c) How is a term produced that contains three factors of x, that is, x3?

By taking x from any three of the factors, in every possible way, and a letter from the remaining factor.

d)  Therefore, what is the coefficient of x3?   a + a + c + d

e)  What is the coefficient of x4?    1 

f)  What is the coefficient of x2?   ab + ac + ad + bc + bd + cd

g)  What is the coefficient of x?   abc + abd + acd + bcd

h)  What is the constant term?   abcd

Problem 4.   Multiply out by taking the correct combinations of the integers.

a)  (x + 1)(x + 3)(x + 4)  = x3 + 8x2 + 19x + 12

b) (x + 1)(x + 2)(x + 3)(x − 1)  = x4 + 5x3 + 5x2 − 5x − 6

Problem 5.   In this multiplication  (x + 1)(x + 2)(x + 3)(x + 4)(x + 5)  what will be the coefficient of x3?   85

Problem 6.   (x + a)5 = (x + a)(x + a)(x + a)(x + a)(x + a)

a)  Upon multiplying out, and before collecting like terms, how many
a)  terms will be produced?   25 = 32

b)  How will a term  x3 a2  be produced?

By taking a from any two factors, in every possible way, and x from the remaining three factors.

c)  How many times will that term be produced?  In other words, upon
c)  adding those like terms, what number will be the coefficient of  x3 a2?

5C2 = 10

Problem 7.   In each row of Pascal's triangle, the sum of the binomial coefficients is 2n.  Why?

2n is the number of terms upon multiplying n binomials. Each binomial coefficient tells how many terms of that kind. Therefore, the sum of all the terms will be 2n.

End of the lessson

Next Topic:  Mathematical induction


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