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# MULTIPLICATION OF SUMS

The binomial coefficients

T gives the coefficients in the product of n equal binomials:

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

If we expanded

(x + b)4,

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

x4, x3b, x2b2, xb3, and b4.

What the binomial theorem does is tell how many terms there are of each kind.  Those are the binomial coefficients.

We will see that determining those coefficients depends on the theory of combinations.  First, then, we will consider the multiplication of any sums; for example,

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

The final product will have this form:

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.

 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).

When we multiply out, we will find six terms.  Each term will contain two factors, namely one letter from each factor:

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

Therefore, to multiply the following,

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

simply write 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.

Now consider these four binomial factors:

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

Let us see how each term in the product will be produced.

If we were actually to multiply out, we would find 24 or 16 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; and finally those multiplied with a binomial will produce 16 terms.

In general:  Multiplication of n binomials produces 2n terms.

Now, in the multiplication of those four binomials, each of those 24 terms will consist of four factors (one from each binomial); hence we expect to find terms such as xbcd,  axcx,  xbxx,  etc.  That is, we will find terms in x,  x2, x3, and x4.

It remains only to determine the coefficients.

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

How then is a term x4 produced?  By taking x from each factor.  xxxx = x4.  There is only one such term.  The coefficient of x4 is 1.

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 ab, c, d  taken 1 at a time:  a + b + c + d.

Terms with x2 will come from taking x from any two factors, in every possible way, and the letters from the remaining two factors:

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

abxx + axcx + axxd + xbcx + xbxd + xxcd

=  (ab + ac + ad + bc + bd + cd)x2

The coefficient of x2 is thus the sum of the combinations of a, b, c, d taken two at a time.

The coefficient of x will be the sum of those combinations taken three at a time:

abcx + abxd + axcd + xbcd   =   (abc + abd + acd + bcd)x

Finally, the term independent of x is formed by those combinations taken four at a time:  abcd.

We have,

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

 =  x4 + (a + b + c + d)x3 + (ab + ac + ad + bc + bd + cd)x2 + (abc + abd + acd + bcd)x + abcd.

Notice again that each term has four factors.

Example.   In this multiplication,

(x + 1)(x − 2)(x + 2)(x + 3),

what number will be the coefficient of x2.

Solution.   Terms containing x2 are formed by taking x from any two factors, in every possible way, and the numbers from the remaining two factors.  Therefore, the coefficient of x2 will be the sum of all possible combinations of the numbers  1, −2, 2, 3,  taken two at a time.  There will be 4C2  or 6  such terms:

 1· −2 + 1· 2 + 1· 3 + −2· 2 + −2· 3 + 2· 3 = −2 + 2 + 3 − 4 − 6 + 6 = −1.

The coefficient of x2 is −1.

The binomial coefficients

Now consider the product of these four equal binomials:

(x + b)(x + b)(x + b)(x + b)

That is, let us expand (x + b)4.

Again, if we were to actually multiply out, then before collecting like terms, we would find 24 or 16  terms; and again every term would consist of four factors (one from each binomial).  There will be terms such as

xxxx,  xxbx,  bbxx,

and so on.  That is, there will be terms in x4, x3b, x2b2, xb3, and b4.

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

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

How will a term x3b be produced?

(x + b)(x + b)(x + b)(x + b)

By taking b from any one factor, and x from the remaining three:

bxxx + xbxx + xxbx + xxxb

Therefore the number of terms equal to x3b will be the number of ways of taking 1 thing -- letter b -- from 4.  There will be 4C1 = 4 terms equal to x3b.  On adding those like terms, we will have 4x3b.

A term x2b2 will come from taking b from any two factors in every possible way, and x from the remaining two.

bbxx + bxbx + bxxb + xbbx + xbxb + xxbb

Therefore, the number of terms equal to x2b2 will be the number of ways of taking 2 things -- letter b -- from 4.   We will have 4C2x2b2, that is, 6x2b2.

A term xb3 will come from taking b from any three factors, and x from the remaining factor; therefore, there will be 4C3 terms equal to xb3; we will have 4xb3.

And finally, a term b4 comes from taking b from each of the four factors; there is only 1 way to do this; we will have 4C4b4 = b4.

This explains the binomial coefficients for the expansion of (x + b)n when n = 4.  They are 1  4  6  4  1.

Again, the binomial coefficients are how many terms there are of each kind.  They are none other than the combinatorial numbers nCk , where k successively takes the values 0, 1, 2, 3, 4.

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

 expansion will have this form: n       (n − k) k an − k bk .

Problem 2.   Imagine multiplying out  (x + y + z)(a + b + 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 + b)(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 + b + 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 + b)5 = (x + b)(x + b)(x + b)(x + b)(x + b)

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

b)  How will a term  x3b2  be produced?

By taking b 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  x3b2?

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.

Next Topic:  Mathematical induction

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