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Mathematics LibreTexts

3.3: The Binomial Theorem

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    60202
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    Here is an algebraic example in which “\(n\) choose \(r\)” arises naturally.

    Example \(\PageIndex{1}\)

    Consider:

    \((a+b)^4 = (a+b)(a+b)(a+b)(a+b)\)

    If you try to multiply this out, you must systematically choose the \(a\) or the \(b\) from each of the four factors, and make sure that you make every possible combination of choices sooner or later.

    One way of breaking this task down into smaller pieces is to separate it into five parts, depending on how many of the factors you choose as from \((4, 3, 2, 1\), or \(0)\). Each time you choose \(4\) of the \(a\)s, you will obtain a single contribution to the coefficient of the term \(a^4\); each time you choose \(3\) of the \(a\)s, you will obtain a single contribution to the term \(a^3b\); each time you choose \(2\) of the \(a\)s, you will obtain a single contribution to the term \(a^2 b^2\); each time you choose \(1\) of the \(a\)s, you will obtain a single contribution to the term \(ab^3\); and each time you choose \(0\) of the \(a\)s, you will obtain a single contribution to the term \(b^4\). In other words, the coefficient of a particular term \(a^i b^{4−i}\) will be the number of ways in which you can choose \(i\) of the factors from which to take an \(a\), taking a \(b\) from the other \(4 − i\) factors (where \(0 ≤ i ≤ 4)\).

    Let’s go through each of these cases separately. By Theorem 3.1.1, there is \( \binom{4}{4} = 1\) way to choose four factors from which to take as. (Clearly, you must choose an a from every one of the four factors.) Thus, the coefficient of \(a^4\) will be \(1\).

    If you want to take as from three of the four factors, Theorem 3.1.1 tells us that there are \( \binom{4}{3} = 4\) ways in which to choose the factors from which you take the \(a\)s. (Specifically, these four ways consist of taking the \(b\) from any one of the four factors, and the as from the other three factors). Thus, the coefficient of \(a^3 b\) will be \(4\).

    If you want to take as from two of the four factors, and bs from the other two, Theorem 3.1.1 tells us that there are \( \binom{4}{2} = 6\) ways in which to choose the factors from which you take the as (then take \(b\)s from the other two factors). This is a small enough example that you could easily work out all six ways by hand if you wish. Thus, the coefficient of \(a^2 b^2\) will be \(6\).

    If you want to take as from one of the four factors, Theorem 3.1.1 tells us that there are \( \binom{4}{1} = 4\) ways in which to choose the factors from which you take the \(a\)s. (Specifically, these four ways consist of taking the \(a\) from any one of the four factors, and the \(b\)s from the other three factors). Thus, the coefficient of \(ab^3\) will be \(4\).

    Finally, by Theorem 3.1.1, there is \( \binom{4}{0} = 1\) way to choose zero factors from which to take \(a\)s. (Clearly, you must choose a \(b\) from every one of the four factors.) Thus, the coefficient of \(b^4\) will be \(1\).

    Putting all of this together, we see that

    \((a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\)

    In fact, if we leave the coefficients in the original form in which we worked them out, we see that

    \((a+b)^4 = \binom{4}{4} a^4 + \binom{4}{3} a^3b + \binom{4}{2} a^2b^2 + \binom{4}{1} ab^3 + \binom{4}{0} b^4\)

    This example generalises into a significant theorem of mathematics:

    Theorem \(\PageIndex{1}\): Binomial Theorem

    For any \(a\) and \(b\), and any natural number \(n\), we have

    \[(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^r b^{n-r} \]

    One special case of this is that

    \[(1+x)^n = \sum_{r=0}^{n} \binom{n}{r} x^r \]

    Proof

    As in Example 2.2.3.1, we see that the coefficient of \(a^r b^{n−r}\) in \((a+b)^n\) will be the number of ways of choosing \(r\) of the \(n\) factors from which we’ll take the \(a\) (taking the \(b\) from the other \(n − r\) factors). By Theorem 2.2.2.1, there are \(\binom{n}{r}\) ways of making this choice. For the special case, begin by observing that \((1 + x)^n = (x+ 1)^n\); then take \(a = x\) and \(b = 1\) in the general formula. Use the fact that \(1^{n−r} = 1\) for any integers \(n\) and \(r\).

    Thus, the values \(\binom{n}{r}\) are the coefficients of the terms in the Binomial Theorem.

    Definition: Binomial Coefficients

    Expressions of the form \(\binom{n}{r}\) are referred to as binomial coefficients.

    There are some nice, simple consequences of the binomial theorem.

    Corollary \(\PageIndex{1}\)

    For any natural number \(n\), we have

    \[\sum_{r=0}^{n} \binom{n}{r} = 2^n\]

    Proof

    This is an immediate consequence of substituting \(a = b = 1\) into the Binomial Theorem.

    Corollary \(\PageIndex{2}\)

    For any natural number \(n\), we have

    \[\sum_{r=0}^{n} \binom{n}{r} = 2^n\]

    Proof

    From the special case of the Binomial Theorem, we have

    \[(1+x)^n = \sum_{r=0}^{n} \binom{n}{r} x^r\]

    If we differentiate both sides, we obtain

    \[n(1+x)^{n-1} = \sum_{r=0}^{n} r \binom{n}{r} x^{r-1}\]

    Substituting \(x = −1\) gives the result (the left-hand side is zero).

    Exercise \(\PageIndex{1}\)

    Use the Binomial Theorem to evaluate the following:

    1) \(\sum_{i=1}^{n} \binom{n}{i}2^i\)

    2) the coefficient of \(a^2 b^3 c^2 d^4\) in \((a + b)^5 (c + d)^6\).

    3) the coefficient of \(a^2 b^6 c^3\) in \((a + b)^5 (b + c)^6\).

    4) the coefficient of \(a^3 b^2\) in \((a + b)^5 + (a + b^2)\).

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