1.9 The binomial theorem

A binomial is an expression with two terms, such as $x + 2$ or $3 - 2x$. Raising a binomial to a small power, such as $(x + 2)^2$, is ordinary algebra, but multiplying out $(x + 2)^{10}$ bracket by bracket would take a long time and invite mistakes. The binomial theorem gives every term of $(a + b)^n$ directly, using a family of whole numbers called binomial coefficients.

The same coefficients count the ways of choosing objects from a set, so the binomial theorem connects algebra with counting and probability. The number of ways of getting exactly 3 heads in 10 coin tosses, for example, is one of these coefficients. Expansions also give quick, accurate approximations to powers of numbers close to 1.

Expanding powers of a binomial

Small powers by multiplying out

The first few powers of $a + b$ can be found by multiplying out one bracket at a time and collecting like terms.

Several patterns appear in every line. The expansion of $(a + b)^n$ has $n + 1$ terms. The powers of $a$ fall by one from $a^n$ to $a^0$, while the powers of $b$ rise by one from $b^0$ to $b^n$, so the two powers in each term always add up to $n$. The coefficients (the numbers in front) read the same forwards and backwards: $1, 4, 6, 4, 1$.

Why the coefficients appear

Expanding $(a + b)^3 = (a + b)(a + b)(a + b)$ means choosing either $a$ or $b$ from each of the three brackets, multiplying the choices, and adding the results for every possible set of choices. The tree diagram lists all $2 \times 2 \times 2 = 8$ ways.

The term $a^2b$ arises whenever $b$ is chosen from exactly one bracket. There are 3 ways to pick that bracket, so the coefficient of $a^2b$ is 3. In general, the coefficient of $a^{n-r}b^r$ in $(a + b)^n$ is the number of ways of choosing which $r$ of the $n$ brackets supply a $b$. This counting idea is the reason behind every result that follows.

Pascal's triangle

Writing the coefficients of $(a + b)^0, (a + b)^1, (a + b)^2, \ldots$ in rows gives Pascal's triangle.

The addition rule works because of the way one row is built from the previous one. Multiplying the expansion of $(a + b)^{n}$ by one more factor $(a + b)$ gives each term of $(a + b)^{n+1}$ two sources: a term multiplied by $a$ and its neighbour multiplied by $b$. Their coefficients add.

Each row also adds up to a power of 2. Substituting $a = b = 1$ into $(a + b)^n$ gives $2^n$, and it also gives the sum of the coefficients in row $n$.

Pascal's triangle is efficient for small powers, but finding row 12 means writing out every row before it. For larger powers, and for finding one term without the others, a formula for each entry is far more useful. That formula uses factorials.

Factorials and binomial coefficients

Factorial notation

The first few values are $1! = 1$, $2! = 2$, $3! = 6$, $4! = 24$, $5! = 120$ and $6! = 720$. Factorials grow very quickly: $10! = 3\,628\,800$. Each factorial is the previous one multiplied by one more number, $n! = n \times (n - 1)!$, and this relation is the key to simplifying expressions with factorials.

The value $0! = 1$ follows from the same relation. Putting $n = 1$ into $n! = n \times (n - 1)!$ gives $1! = 1 \times 0!$, so $0! = 1$. It also makes the formulas for binomial coefficients work at the ends of each row.

The binomial coefficient

The formula comes from counting. Choosing $r$ objects one after another from $n$ can be done in $n(n - 1)\cdots(n - r + 1) = \frac{n!}{(n - r)!}$ ordered ways, but each group of $r$ objects is counted $r!$ times, once for every order in which it could have been picked. Dividing by $r!$ removes the repeats. Counting arguments of this kind are developed fully in 1.10 (HL).

Some values are worth recognising at once.

Evaluating binomial coefficients by hand

Most of the factorials in $\frac{n!}{r!(n - r)!}$ cancel. Cancelling the larger of $r!$ and $(n - r)!$ leaves a short product on the top divided by a small factorial:

The top has exactly $r$ factors, counting down from $n$.

Evaluating binomial coefficients with technology

A graphic display calculator (GDC) has an nCr function in its probability menu: entering $n$, then nCr, then $r$ gives $\binom{n}{r}$ directly. A GDC can also display a whole row of Pascal's triangle by putting a list of values $r = 0, 1, \ldots, n$ into nCr, which is useful for checking an expansion. Spreadsheets have the same function (COMBIN in most programs).

Symmetry and other properties

Each row of Pascal's triangle reads the same in both directions. Choosing $r$ objects to take is the same as choosing the $n - r$ objects to leave behind, so

The formula agrees: swapping $r$ and $n - r$ in $\frac{n!}{r!(n - r)!}$ only changes the order of the two factorials on the bottom. Symmetry halves the work by hand, as in $\binom{20}{17} = \binom{20}{3}$.

The addition rule of Pascal's triangle can also be written with binomial coefficients:

For example, $\binom{5}{2} + \binom{5}{3} = 10 + 10 = 20 = \binom{6}{3}$, the shaded entries in Pascal's triangle.

The binomial theorem

Statement of the theorem

Combining the counting argument with the formula for $\binom{n}{r}$ gives the binomial theorem.

The theorem holds for any $a$ and $b$: numbers, variables, negative terms, fractions or powers of $x$. The reason is the one shown by the tree diagram. In the product of $n$ brackets $(a + b)$, the term $a^{n-r}b^r$ arises once for every way of choosing which $r$ brackets supply the $b$, and there are $\binom{n}{r}$ such ways.

Expanding with the binomial theorem

To expand a binomial such as $(2x - 3)^4$, identify $a$ and $b$ with their signs and coefficients, here $a = 2x$ and $b = -3$, and substitute them as whole bracketed terms. Laying the work out as a table keeps the powers and signs under control.

Terms with negative powers of x

When one term of the binomial is a reciprocal such as $\frac{2}{x}$, the powers of $x$ from the two parts combine. In $(x - \frac{2}{x})^5$ the term with $r$ factors of $-\frac{2}{x}$ contains $x^{5-r} \times x^{-r} = x^{5 - 2r}$, so the powers of $x$ fall by 2 each time.

The general term

Finding one term without expanding

Often only one term of an expansion is needed, such as the coefficient of $x^3$ in $(3 + 2x)^7$. Each term of the binomial theorem has the same shape, so a single formula describes all of them.

The method for finding a particular term has three steps.

1. Write the general term with $a$ and $b$ substituted, and collect the powers of $x$ into a single power $x^{\text{something in } r}$.

2. Set that power equal to the power wanted and solve for $r$. The value of $r$ must be an integer from $0$ to $n$; if it is not, there is no such term and the coefficient is 0.

3. Substitute $r$ back into the general term and evaluate.

The constant term

The constant term of an expansion is the term independent of $x$, that is, the term in $x^0$. It exists only when the powers of $x$ from $a$ and $b$ can cancel exactly, which happens when one part of the binomial has a positive power of $x$ and the other a negative power.

The same method finds any power. A useful check comes before any arithmetic: if the equation for $r$ gives a fraction or a value outside $0 \le r \le n$, the term does not exist. In $\left(2x - \frac{1}{x}\right)^6$ every power of $x$ is even, because $6 - 2r$ is even, so there is no term in $x^3$ and its coefficient is 0.

Finding an unknown constant

When a binomial contains an unknown constant, a given coefficient produces an equation for it. Write the relevant term in terms of the constant, set it equal to the given value and solve.

Products of a bracket and a binomial expansion

Some questions ask for a coefficient in a product such as $(2 - x)(1 + 3x)^5$. There is no need to expand the whole product. A term in $x^2$ comes from multiplying each term of the first bracket by the term of the expansion that brings the power up to exactly $x^2$:

• the $2$ in the first bracket times the $x^2$ term of $(1 + 3x)^5$;

• the $-x$ in the first bracket times the $x$ term of $(1 + 3x)^5$.

So only the first few terms of the binomial expansion are needed.

The first bracket can itself contain powers of $x$ that skip a term. In $(1 - x^2)(2 + x)^5$, the $x^3$ term needs the $x^3$ term of $(2 + x)^5$ (times 1) and the $x$ term (times $-x^2$), but not the $x^2$ term.

Approximations using the binomial theorem

Powers of numbers close to 1

When $x$ is small, the powers $x^2, x^3, \ldots$ are much smaller still. In the expansion of $(1 + x)^n$ the later terms shrink rapidly, so the first few terms give a good approximation. This is the binomial theorem used as a tool for estimation.

The graph compares $y = (1 + x)^5$ with the first two and the first three terms of its expansion. Close to $x = 0$ the three curves are almost indistinguishable; further away, the line $y = 1 + 5x$ separates first and the quadratic follows the curve for longer.

Choosing the value of x

The approximation works only when the number substituted for $x$ is small. To estimate $0.995^8$, write $0.995 = 1 - 0.005$; to estimate $1.99^6$, write $1.99 = 2 - 0.01$ and use the expansion of $(2 - x)^6$. A question often gives the expansion in terms of $x$ first and asks for a suitable value of $x$.

Binomial coefficients and probability

The binomial coefficients also count outcomes. When a fair coin is tossed $n$ times, each sequence of heads and tails is equally likely, and there are $2^n$ sequences in total. A sequence with exactly $r$ heads is fixed by choosing which $r$ of the $n$ tosses are heads, so there are $\binom{n}{r}$ such sequences. This is the same choice as picking which brackets supply $b$ in $(a + b)^n$, and it explains why row $n$ of Pascal's triangle adds up to $2^n$.

Expanding $\left(\frac{1}{2} + \frac{1}{2}\right)^6$ with the binomial theorem gives one term for each possible number of heads, and the term with $r = 3$ is $\binom{6}{3}\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^3 = \frac{20}{64}$. Replacing $\frac{1}{2}$ by probabilities $p$ and $1 - p$ leads to the binomial distribution in 4.8.

Exam-style questions on the binomial theorem

Binomial questions appear on both papers. The non-calculator paper tests expansions and coefficients that need careful arithmetic by hand; the calculator paper adds unknown constants and equations that are faster to solve with a GDC.

A second common type gives a coefficient and asks for an unknown constant, then uses that constant in a further part.

Self-check

• To do:

  I can evaluate factorials and simplify expressions such as $\frac{(n+1)!}{(n-1)!}$.

• To do:

  I can evaluate $\binom{n}{r}$ by hand, with a GDC and from Pascal's triangle, and use $\binom{n}{r} = \binom{n}{n-r}$.

• To do:

  I can expand $(a + b)^n$ when $a$ or $b$ is negative, a multiple of $x$ or a reciprocal of $x$

• To do:

  I can write the general term $\binom{n}{r}a^{n-r}b^r$ and use it to find a given term, coefficient or the constant term.

• To do:

  I can explain why an expansion has no term in a particular power of $x$.

• To do:

  I can find an unknown constant from a given coefficient, keeping all valid solutions.

• To do:

  I can find a coefficient in the product of a bracket and a binomial expansion.

• To do:

  I can choose a suitable value of $x$ and use the first terms of an expansion to estimate a power such as $1.02^{10}$.

1. (a)

   Show that $\displaystyle\int_0^1 f_n(x)\,\mathrm{d}x = 1$ for all $n \in \mathbb{Z}^+$.

   [2]2 marks

   Markscheme (a)

   $\int_0^1 n(1 - x)^{n - 1}\,\mathrm{d}x = \left[-(1 - x)^n\right]_0^1$

2. (b)(i)

   Consider the case $n = 3$.

   Find $\mathrm{E}(X_3)$.

   [2]2 marks

   Markscheme (b)(i)

   $\mathrm{E}(X_3) = \int_0^1 3x(1 - x)^2\,\mathrm{d}x$

3. (b)(ii)

   Find the median of $X_3$, and hence state whether the median of $X_3$ is greater than or less than $\mathrm{E}(X_3)$

4. (c)

   By using the substitution $u = 1 - x$, show that $\mathrm{E}(X_n) = \dfrac{1}{n + 1}$

5. (d)

   Show that $\mathrm{Var}(X_n) = \dfrac{n}{(n + 1)^2(n + 2)}$

6. (e)

   Show that the median of $X_n$ is $m_n = 1 - 2^{-\frac{1}{n}}$

7. (f)(i)

   Show that, for $n \in \mathbb{Z}^+$, the inequality $\mathrm{E}(X_n) > m_n$

8. (f)(ii)

   By using the binomial theorem, prove that $\mathrm{E}(X_n) > m_n$ for all integers $n \ge 2$.

   [3]

9. (g)(i)

   Use l'Hôpital's rule to find $\displaystyle\lim_{n \to \infty} n\,m_n$.

   [3]3 marks

   Markscheme (g)(i)

   $n\,m_n = \frac{1 - 2^{-\frac{1}{n}}}{\frac{1}{n}}$

10. (g)(ii)

    Hence find $\displaystyle\lim_{n \to \infty} \frac{m_n}{\mathrm{E}(X_n)}$

(b) Write the larger factorial in terms of the smaller one: $(n + 1)! = (n + 1) \times n \times (n - 1)!$.

Cancelling first is quicker: $\frac{8}{4 \times 2} = 1$ and $\frac{6}{3} = 2$, leaving $9 \times 7 \times 2 = 126$.

(b) Choosing 17 objects from 20 is the same as choosing the 3 to leave out, so $\binom{20}{17} = \binom{20}{3}$.

So $n = 8$ or $n = -7$. Since $n$ must be a positive integer, $n = 8$.

Check: $\binom{8}{2} = \frac{8 \times 7}{2} = 28$.

Set $18 - 3r = 6$, so $r = 4$, which is an integer between 0 and 9.

The term in $x^6$ is $10\,206x^6$.

Pair them with the terms of $1 - x^2$.

The coefficient of $x^3$ is $-40$.

(b) Choose $x$ so that $1 - \frac{x}{2} = 0.995$: then $\frac{x}{2} = 0.005$ and $x = 0.01$.

A calculator gives $0.995^8 = 0.960693\ldots$, which also rounds to $0.9607$.

One mark is usually for the correct binomial coefficients, one for powers of 2 and $-x$ in the right places, and one for the simplified terms.

(b) "Hence" means use part (a). The $x^2$ term comes from $1 \times 240x^2$ and $3x \times (-192x)$.

The coefficient of $x^2$ is $-336$.

(c) Choose $x$ so that $2 - x = 1.99$, which gives $x = 0.01$.

So $1.99^6 \approx 62.104$. (A calculator gives $62.10384\ldots$, which agrees to 3 decimal places.)

(b) The constant term has $9 - 3r = 0$, so $r = 3$.

(c) For $x^3$, $9 - 3r = 3$ gives $r = 2$.

The coefficient of $x^3$ is 144.

(d) A term in $x^2$ would need $9 - 3r = 2$, so $r = \frac{7}{3}$, which is not an integer. Every power of $x$ in the expansion is of the form $9 - 3r$, a multiple of 3, so there is no term in $x^2$.

$= 0 - (-1) = 1$ AG

$= \frac{1}{4}$ A1

[2]2 marks

[4]4 marks

[5]5 marks

[2]2 marks

[3]3 marks

Note: Accept the substitution $h = \frac{1}{n}$, giving $\lim_{h \to 0^+} \frac{1 - 2^{-h}}{h}$.

differentiating numerator and denominator: $\frac{-2^{-\frac{1}{n}}\ln 2 \times \frac{1}{n^2}}{-\frac{1}{n^2}} = 2^{-\frac{1}{n}}\ln 2$ A1

$\lim_{n \to \infty} n\,m_n = \ln 2$ A1

[2]2 marks

[2]2 marks