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26

# MULTIPLICATION OF SUMS

## A proof of the binomial theorem

T gives the coefficient of each term 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,

we would find terms in

x4, x3a, x2a2, xa3, and a4.

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

1  4  6  4  1

There is but 1 term in x4; 4 in x3a; 6 in x2a2; 4 in xa3; 1 in a4.

The binomial theorem is that those coefficients are the combinatorial numbers.

To prove that, we will first consider the multiplication of any sums; for example:

(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.

Example.   (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.

How is the coefficient of each power of x produced?

Each term in the product will have 4 factors, one from each binomial.

x4 comes from taking x from each binomial.  xxxx = x4.  There is only one such term.  The coefficient of x4 is 1.

Terms with x3 are formed by taking  x  from three binomials, in every possible way, and the letter from the remaining one.

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 or 4:  The number of combinations of 4 letters taken 1 at a time.

Next, terms with x2 will come from taking x from two binomials, in every possible way, and the letters from the remaining two. There will be 4C2 or 6 such terms:

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

4C2 is the number of ways of selecting 2 letters from 4: the 4 binomials.

A term in x will be produced by taking x from 1 binomial 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 the 4 binomials.

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

The binomial theorem

If, in the four binomials above, we make all the letters equal, then we have (x + a)4.

Each coefficient of a power of x will be a sum of powers of the a's.

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

 =  x4 + (a + a + a + a)x3 + (a² + a² + a² + a² + a² + a²)x² + (a³ + a³ + a³ + a³)x + a4.

Now, the binomial coefficients are how many terms of each kind.

We saw that the number of terms with x4 is 4C0 or 1. 4C0 is the coefficient of x4.

The number of terms with x³ is 4C1. That is the number of a's in the coefficient of x3. The coefficient of ax³ is 4C1.

The number of a²s in the coefficient of x2 is 4C2. The coefficient of a²x² is 4C2.

The number of a³s in the coefficient of x is 4C3. The coefficient of a³x is 4C3.

The number of a4s in the constant term is 4C4. That is the coefficient of a4.

In other words:

 (x + a)4 = 4C0x4 + 4C1ax³ + 4C2a²x² + 4C3a³x + 4C4a4 = x4 + 4ax³ + 6a²x² + 4a³x + a4.

The binomial coefficients are the combinatorial numbers.

This can be generalized for any exponent n.  The binomial theorem states that in the expansion of (x + a)n, the coefficients are the combinatorial numbers nCk , where k—the exponent of a—successively takes the values 0, 1, 2, . . . , n.

 Each term in the expansion will have this form: n (n − k) k an − kbk .

Compare Example 4, Lesson 25.

Problem 1.   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 2.   Write the product by taking the correct combinations of the integers.

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

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

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

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

Problem 4.

(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  x3a2  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  x3a2?

The number of ways of choosing 2 things—letter a—from 5:

5C2 = 10

Problem 5.   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 is the number of terms of that kind. Therefore, the sum of all the terms will be 2n. Next Topic:  Mathematical induction

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