The general term

What you will be able to do

• write the general term $T_{r+1}$ of any binomial expansion;
• find a stated term, counting from the start or from the end;
• find the term that contains a given power of $x$, and its coefficient;
• decide when no such term exists.

One formula for every term

The binomial theorem lists the terms in order. The term with $r$ copies of $b$ is

It is called $T_{r+1}$ because the first term has $r = 0$. So the $5$th term has $r = 4$, and the $10$th term has $r = 9$. The number of the term is always one more than $r$.

The formula gives a term without writing the ones before it. In $(a + b)^9$, the $3$rd term is ${}^9C_2\,a^7b^2 = 36a^7b^2$.

Counting from the end

The expansion of $(a + b)^n$ has $n + 1$ terms. The last term is the $(n + 1)$th, so the $p$th term from the end is the $(n - p + 2)$th term from the start. For example, the $3$rd term from the end of $(a + b)^{10}$ is $T_9$, which is ${}^{10}C_8\,a^2b^8$.

Finding the term with a given power of $x$

When $a$ or $b$ holds a power of $x$, the power of $x$ changes from term to term. Follow four steps.

1. Write the general term. Put the two parts of the bracket into $T_{r+1} = {}^nC_r\,a^{n-r}b^r$.
2. Collect the powers of $x$. Use $(x^p)^m = x^{pm}$ and $\left(\dfrac{1}{x}\right)^r = x^{-r}$. The power of $x$ is now an expression in $r$.
3. Match the power. Set the expression equal to the power you want and solve for $r$.
4. Check $r$. It must be a whole number, from $0$ to $n$. If it is not, there is no such term, and the coefficient is $0$.

A worked example

In the expansion of $\left(2x^2 - \dfrac{1}{x}\right)^6$, find the coefficient of $x^3$ and the coefficient of $x^4$.

1. The general term. $T_{r+1} = {}^6C_r\,(2x^2)^{6-r}\left(-\dfrac{1}{x}\right)^r = {}^6C_r\,2^{6-r}(-1)^r\,x^{12-2r-r}$.
2. The power of $x$. It is $12 - 3r$.
3. For $x^3$. $12 - 3r = 3$ gives $r = 3$, a whole number between $0$ and $6$. The coefficient is ${}^6C_3 \times 2^3 \times (-1)^3 = 20 \times 8 \times (-1) = -160$.
4. For $x^4$. $12 - 3r = 4$ gives $r = \dfrac{8}{3}$, which is not a whole number. So there is no term with $x^4$, and its coefficient is $0$.

Practise with new numbers

Try it in Python

The program lists every term of $\left(x^2 + \dfrac{3}{x}\right)^7$ with its coefficient and its power of $x$. It marks the term with the power you ask for.

Common mistakes

• Mixing up $T_r$ and $T_{r+1}$. The formula ${}^nC_r\,a^{n-r}b^r$ is the $(r + 1)$th term.
• Raising only the letter to a power. In $(2x^2)^{6-r}$ the number $2$ is raised to the power $6 - r$ too.
• Mishandling $\dfrac{1}{x}$. $\left(\dfrac{1}{x}\right)^r = x^{-r}$, so the power of $x$ falls by $r$.
• Skipping the check on $r$. A fraction, or a value above $n$, means there is no such term.

Remember these

• $T_{r+1} = {}^nC_r\,a^{n-r}b^r$, and the number of the term is $r + 1$.
• The $p$th term from the end is the $(n - p + 2)$th term from the start.
• For a given power of $x$: write the general term, match the power, solve for $r$, then check $r$.
• If $r$ is not a whole number from $0$ to $n$, there is no such term.

Check yourself

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

1. Find the $4$th term of the expansion of $(m + n)^{11}$.

2. In $(x + y)^{12}$, which term contains $x^9$?

3. Find the $5$th term of $(2 - x)^8$.

4. What is the power of $x$ in the $(r + 1)$th term of $\left(x^3 + \dfrac{1}{x}\right)^n$?

5. Does the expansion of $\left(x^2 + \dfrac{1}{x}\right)^9$ have a term with $x^5$?