Pascal's triangle and the expansion

What you will be able to do

• expand a small power of a sum with Pascal's triangle, and any power with the binomial theorem;
• say why the coefficient of $a^{n-r}b^r$ is ${}^nC_r$;
• expand $(a - b)^n$ and $(1 + x)^n$, and count the terms;
• add up the coefficients of an expansion without writing it out.

Squares and cubes of a sum

You know that $(a + b)^2 = a^2 + 2ab + b^2$. Multiply by $a + b$ once more and you get the cube:

One more multiplication gives

Look at the pattern in these expansions.

• The power of $a$ goes down by one in each term, and the power of $b$ goes up by one.
• In every term the two powers add up to $n$.
• There are $n + 1$ terms.
• The numbers in front are called coefficients. They are $1, 2, 1$ for $n = 2$, then $1, 3, 3, 1$, then $1, 4, 6, 4, 1$.

Pascal's triangle

Write the coefficients as rows, one row for each power. The rows form a triangle named after the French mathematician Blaise Pascal.

Each number is the sum of the two numbers just above it. Row $6$ is $1, 6, 15, 20, 15, 6, 1$, because $5 + 10 = 15$ and $10 + 10 = 20$. Every row starts and ends with $1$, and every row reads the same forwards and backwards.

Rows are counted from $n = 0$, which is the single number $1$, because $(a + b)^0 = 1$. So row $n$ holds the coefficients of $(a + b)^n$. For example, row $5$ gives

Pascal's triangle is quick for small powers. For $(a + b)^{20}$ you would need twenty rows. We need a formula that gives any coefficient at once.

Why the coefficient is a choice

Think of $(a + b)^3$ as three brackets multiplied together:

To make one product you take either $a$ or $b$ from each bracket. That gives $2 \times 2 \times 2 = 8$ products, such as $aab$ and $bab$.

Group the products by the number of $b$'s. A product with two $b$'s comes from choosing which two of the three brackets give a $b$. That can be done in ${}^3C_2 = 3$ ways, so the term $ab^2$ has coefficient $3$.

The same idea works for every $n$. In $(a + b)^n$, a product with $r$ letters $b$ and $n - r$ letters $a$ comes from choosing $r$ of the $n$ brackets for the $b$. There are ${}^nC_r$ ways, so the coefficient of $a^{n-r}b^r$ is ${}^nC_r$.

The binomial theorem

The binomial theorem says that for every whole number $n$,

In sigma notation this is $\displaystyle\sum_{r=0}^{n} {}^nC_r\,a^{n-r}b^r$. Row $n$ of Pascal's triangle is ${}^nC_0, {}^nC_1, \dots, {}^nC_n$. Since ${}^nC_r = {}^nC_{n-r}$, each row reads the same in both directions.

Three versions are used all the time.

• A difference. Replace $b$ by $-b$ to get $(a - b)^n = \displaystyle\sum_{r=0}^{n} (-1)^r\,{}^nC_r\,a^{n-r}b^r$. The signs alternate: $+, -, +, -, \dots$
• A power of $1 + x$. Put $a = 1$ and $b = x$ to get $(1 + x)^n = \displaystyle\sum_{r=0}^{n} {}^nC_r\,x^r$.
• The sum of a row. Put $a = 1$ and $b = 1$ to get ${}^nC_0 + {}^nC_1 + \dots + {}^nC_n = 2^n$.

A worked example

Expand $(x - 2y)^5$, and check the answer by adding the coefficients.

1. Choose $a$ and $b$. Take $a = x$ and $b = -2y$. The term with $r$ copies of $b$ is ${}^5C_r\,x^{5-r}(-2y)^r$.
2. Read row 5. The coefficients are $1, 5, 10, 10, 5, 1$.
3. Write each term. $x^5$, then $5x^4(-2y) = -10x^4y$, then $10x^3(4y^2) = 40x^3y^2$, then $10x^2(-8y^3) = -80x^2y^3$, then $5x(16y^4) = 80xy^4$, and last $-32y^5$.
4. Check. Put $x = 1$ and $y = 1$ in $(x - 2y)^5$. The result is $(1 - 2)^5 = -1$. The coefficients add up to $1 - 10 + 40 - 80 + 80 - 32 = -1$.

Practise with new numbers

Try it in Python

The program multiplies out $(1 + x)^6$ one bracket at a time. After each bracket it prints the list of coefficients. Then it compares the last list with the formula.

Common mistakes

• Forgetting the numbers inside the brackets. In $(2x + 3)^n$ the coefficient of each term is ${}^nC_r\,2^{n-r}3^r$, not just ${}^nC_r$.
• Losing the signs in $(a - b)^n$. Every odd power of $b$ carries a minus sign.
• Counting $n$ terms. There are $n + 1$ terms, one for each $r$ from $0$ to $n$.
• Reading the wrong row. Row $0$ is $(a + b)^0$, so $(a + b)^n$ is row $n$, not row $n + 1$.

Remember these

• $(a + b)^n = \displaystyle\sum_{r=0}^{n} {}^nC_r\,a^{n-r}b^r$, with $n + 1$ terms.
• The coefficients are row $n$ of Pascal's triangle. Each number is the sum of the two above it.
• ${}^nC_r$ counts the ways to choose which $r$ of the $n$ brackets give a $b$.
• For $(a - b)^n$ the signs alternate. Each row adds up to $2^n$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. How many terms does the expansion of $(a + b)^{12}$ have?

2. Find the coefficient of $a^2b^4$ in the expansion of $(a + b)^6$.

3. Expand $(1 - 2x)^3$.

4. What is the sum of the coefficients in the expansion of $(3x - 1)^4$?

5. What do the numbers in row $7$ of Pascal's triangle add up to?