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The binomial theorem

Counting and combinations · level 7, lesson 2 of 6

A question to start

Work out the powers of 11:

\[11^1 = 11, \qquad 11^2 = 121, \qquad 11^3 = 1331, \qquad 11^4 = 14641.\]

The digits are 1 1, then 1 2 1, then 1 3 3 1, then 1 4 6 4 1: the rows of Pascal's triangle. And \(101^3 = 1030301\) hides row 3 again, with zeros in between. Why should multiplying by 11 or 101 know anything about choosing teams? And what goes wrong at \(11^5 = 161051\), where the pattern seems to break? This lesson answers both, and gives you a way to find any single term of an expansion like \((2x - 3)^8\) in one line.

The idea

Sigma: a short way to write a long sum

A binomial is an expression with two terms, such as \(a + b\) or \(2x - 3\). In the last lesson you found that \((a + b)^n\) is a sum of \(n + 1\) terms, one for each number of \(b\)'s from 0 to \(n\). Writing them all out with dots is clumsy, so mathematicians use the Greek capital letter sigma, \(\Sigma\), which means "sum":

\[\sum_{k=0}^{3} k^2 = 0^2 + 1^2 + 2^2 + 3^2 = 14.\]

Read it as: "let \(k\) run through \(0, 1, 2, 3\), work out \(k^2\) each time, and add the results". The letter \(k\) is a counter: it has no value of its own, and any other letter would do just as well.

The theorem

The binomial theorem. For any whole number \(n \ge 0\) and any numbers \(a\) and \(b\),

\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{\,n-k} b^{\,k} = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + b^n.\]

The numbers \(\dbinom{n}{k}\) are called binomial coefficients because of this theorem, and they are exactly the entries of row \(n\) of Pascal's triangle.

1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 row 8 row 0

Each term has three parts: a coefficient \(\dbinom{n}{k}\), a power of \(a\) that goes down from \(n\) to 0, and a power of \(b\) that goes up from 0 to \(n\). In every term the two powers add to \(n\). The term with \(b^k\) is often called the general term,

\[T_{k+1} = \binom{n}{k} a^{\,n-k} b^{\,k},\]

numbered \(k + 1\) because the first term has \(k = 0\).

Proof 1: by choosing

This is the argument of lesson 1, now written as a proof. Expanding \((a + b)(a + b)\cdots(a + b)\) with \(n\) brackets means choosing \(a\) or \(b\) from each bracket in every possible way and adding the products. A choice that takes \(b\) from exactly \(k\) brackets gives \(a^{n-k}b^k\). Such a choice is fixed by the set of \(k\) brackets that gave \(b\), and there are \(\dbinom{n}{k}\) such sets. So \(a^{n-k}b^k\) appears exactly \(\dbinom{n}{k}\) times, for each \(k\) from 0 to \(n\). ∎

Proof 2: by induction, using Pascal's rule

A second proof shows how the theorem and Pascal's triangle fit together. Induction proves a statement for every \(n\) in two moves: show it is true for the first \(n\) (the base case), then show that if it is true for some \(n\) it must be true for \(n + 1\) (the step). Then it is true for \(n = 0\), so for 1, so for 2, and on forever, like a row of dominoes.

Base case. For \(n = 0\): \((a + b)^0 = 1\), and the sum has one term, \(\dbinom{0}{0}a^0b^0 = 1\). ✓

Step. Suppose the theorem is true for some \(n\). Then

\[(a + b)^{n+1} = (a + b)(a + b)^n = a\,(a + b)^n + b\,(a + b)^n.\]

Using the theorem for \(n\), the term \(a^{n+1-k}b^k\) can come from two places: \(a\) times the term \(\dbinom{n}{k}a^{n-k}b^k\), or \(b\) times the term \(\dbinom{n}{k-1}a^{n-k+1}b^{k-1}\). So its coefficient is

\[\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k},\]

by Pascal's rule from level 5: each entry is the sum of the two above it. That is exactly the coefficient the theorem promises for \(n + 1\). ∎

The two proofs say the same thing from two sides. Multiplying by one more bracket \((a + b)\) is exactly what building one more row of Pascal's triangle does.

Signs: when \(b\) is negative

For \((a - b)^n\), put \(-b\) in place of \(b\). The term with \((-b)^k\) has the sign of \((-1)^k\): plus when \(k\) is even, minus when \(k\) is odd. So the signs alternate:

\[(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3.\]

Finding one term without the rest

Often you want just one term, say the \(x^5\) term of \((2x - 3)^8\). Expanding all nine terms is slow. Instead:

  1. Name the two parts carefully, with their signs: \(a = 2x\) and \(b = -3\), \(n = 8\).
  2. Write the general term: \(\dbinom{8}{k}(2x)^{8-k}(-3)^k\).
  3. Choose \(k\) to get the power you want: \(x^5\) needs \(8 - k = 5\), so \(k = 3\).
  4. Work it out:
\[\binom{8}{3}(2x)^5(-3)^3 = 56 \times 32x^5 \times (-27) = -48\,384\,x^5.\]

The \(x^5\) term is \(-48\,384x^5\), and its coefficient is \(-48\,384\).

The powers of 11, explained

\(11^n = (10 + 1)^n\). With \(a = 10\) and \(b = 1\), the term \(\dbinom{n}{k}10^{n-k}\) puts the coefficient \(\dbinom{n}{k}\) in the \(10^{n-k}\) column, like a digit:

\[11^4 = 1 \cdot 10^4 + 4 \cdot 10^3 + 6 \cdot 10^2 + 4 \cdot 10 + 1 = 14\,641.\]

In row 5 the coefficients are 1, 5, 10, 10, 5, 1. Two of them are not single digits, so they carry into the next column, and 1 5 10 10 5 1 becomes 161051. The pattern never broke; it only got hidden by carrying. With \(101 = 100 + 1\) each coefficient gets two columns of room, which is why \(101^3 = 1\,030\,301\) shows 1, 3, 3, 1 with zeros between.

Quick estimates

When \(x\) is small, the powers \(x^2, x^3, \ldots\) are tiny, so the first few terms of \((1 + x)^n\) are almost the whole answer:

\[(1 + x)^n \approx 1 + nx + \binom{n}{2}x^2.\]

For example \(1.02^8 \approx 1 + 8(0.02) + 28(0.0004) = 1.1712\). The true value is \(1.17166\ldots\) This is how interest on money was estimated before calculators: 2% a year for 8 years grows money by about 17%.

Worked examples

1. Expand \((2x - 1)^5\). Here \(a = 2x\), \(b = -1\), \(n = 5\), and row 5 is 1, 5, 10, 10, 5, 1.

\[ \begin{aligned} (2x - 1)^5 &= (2x)^5 - 5(2x)^4 + 10(2x)^3 - 10(2x)^2 + 5(2x) - 1\\ &= 32x^5 - 80x^4 + 80x^3 - 40x^2 + 10x - 1. \end{aligned} \]

Check with \(x = 1\): the left side is \(1^5 = 1\), the right side is \(32 - 80 + 80 - 40 + 10 - 1 = 1\). ✓

2. The \(x^5\) term of \((2x - 3)^8\). Done above: \(k = 3\) gives \(\dbinom{8}{3}(2x)^5(-3)^3 = -48\,384x^5\). A sign check: \((-3)^3\) is negative and everything else is positive, so the term is negative. ✓

3. The term with no \(x\). Find the constant term of \(\left(x^2 - \dfrac{1}{x}\right)^9\).

  • General term: \(\dbinom{9}{k}(x^2)^{9-k}\left(-\dfrac{1}{x}\right)^k = \dbinom{9}{k}(-1)^k x^{18 - 2k - k} = \dbinom{9}{k}(-1)^k x^{18 - 3k}\).
  • No \(x\) means \(18 - 3k = 0\), so \(k = 6\).
  • The term is \(\dbinom{9}{6}(-1)^6 = 84\).

4. Working backwards. In \((1 + x)^n\) the coefficient of \(x^2\) is 45. Find \(n\).

The coefficient is \(\dbinom{n}{2} = \dfrac{n(n-1)}{2}\). So \(n(n - 1) = 90 = 10 \times 9\), and \(n = 10\).

5. An estimate. Estimate \(1.02^8\) to four decimal places using three terms, then four. Three terms: \(1 + 0.16 + 0.0112 = 1.1712\). The fourth term is \(\dbinom{8}{3}(0.02)^3 = 56 \times 0.000008 = 0.000448\), giving \(1.171648\). The true value is \(1.1716594\ldots\), so four terms are right to five decimal places.

Try it

  1. Write \((a + b)^6\) with \(\Sigma\), then expand it fully.
  2. Expand \((x - 2)^4\), and check by putting \(x = 1\).
  3. Find the coefficient of \(x^3\) in \((3 + x)^7\).
  4. Find the coefficient of \(x^4\) in \((1 - 2x)^9\).
  5. In \(\left(2x^2 + \dfrac{1}{x}\right)^6\), find the \(x^6\) term and the constant term.
  6. Estimate \(0.98^{10}\) by writing it as \((1 - 0.02)^{10}\) and using four terms.
  7. The coefficient of \(x^2\) in \((1 + x)^n\) is 66. Find \(n\).
  8. Show that \((\sqrt{2} + 1)^4 + (\sqrt{2} - 1)^4\) is a whole number, and find it. (Hint: what happens to the odd powers of 1?)
Answers
  1. \((a + b)^6 = \sum_{k=0}^{6} \binom{6}{k} a^{6-k} b^k = a^6 + 6a^5b + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6ab^5 + b^6\).
  2. \(a = x\), \(b = -2\), row 4 is 1, 4, 6, 4, 1: \(x^4 + 4x^3(-2) + 6x^2(4) + 4x(-8) + 16 = x^4 - 8x^3 + 24x^2 - 32x + 16\). Check: \((1 - 2)^4 = 1\) and \(1 - 8 + 24 - 32 + 16 = 1\). ✓
  3. \(a = 3\), \(b = x\): the term with \(x^3\) has \(k = 3\): \(\binom{7}{3}3^4x^3 = 35 \times 81x^3\), so the coefficient is 2835.
  4. \(a = 1\), \(b = -2x\), \(k = 4\): \(\binom{9}{4}(-2x)^4 = 126 \times 16x^4\), so the coefficient is 2016 (positive, since the power 4 is even).
  5. General term \(\binom{6}{k}(2x^2)^{6-k}x^{-k} = \binom{6}{k}2^{6-k}x^{12 - 3k}\). For \(x^6\): \(12 - 3k = 6\), \(k = 2\): \(15 \times 2^4 = 240\), so the term is \(240x^6\). Constant: \(12 - 3k = 0\), \(k = 4\): \(15 \times 2^2 = 60\).
  6. \(1 - 10(0.02) + 45(0.0004) - 120(0.000008) = 1 - 0.2 + 0.018 - 0.00096 = 0.81704\). (True value \(0.81707\ldots\))
  7. \(\frac{n(n-1)}{2} = 66\), so \(n(n - 1) = 132 = 12 \times 11\), and \(n = 12\).
  8. Expand both with \(a = \sqrt{2}\), \(b = \pm 1\). The terms with odd powers of \(b\) have opposite signs and cancel; the even ones double: \(2\left[(\sqrt{2})^4 + 6(\sqrt{2})^2 + 1\right] = 2(4 + 12 + 1) = 34\). A calculator agrees: \((\sqrt{2} + 1)^4 \approx 33.97\) and \((\sqrt{2} - 1)^4 \approx 0.03\), adding to 34.

Common mistake

Not raising the whole part to the power. In \((2x - 3)^8\) the part \(a\) is \(2x\), so the term contains \((2x)^5 = 32x^5\), not \(2x^5\); and \(b\) is \(-3\), so the sign comes from \((-3)^3\). Put brackets round each part before you raise it to a power.

Using \(k\) as the power of \(a\). In the form \(\dbinom{n}{k}a^{n-k}b^k\), \(k\) is the power of the second part. If you mix them up you will usually still get the right number (because \(\dbinom{n}{k} = \dbinom{n}{n-k}\)) but the wrong power of the numbers inside, such as \(2^3\) instead of \(2^5\).

Where it leads