A proof of the binomial theorem - Topics in precalculus

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MULTIPLICATION OF SUMS

A proof of the binomial theorem

TO PROVE THE BINOMIAL THEOREM we will first 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.

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	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 factor.

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 factor); 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.

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.

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.

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

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 theorem

Now consider the product of these four equal factors:

(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). 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.

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

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 ⁿC_k , where k successively takes on the value 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 ⁿC_k , where k successively takes on the value 0, 1, 2, . . . , n. Each term in the