The binomial expansion for positive integer powers

The binomial theorem expands a power of a two-term expression without repeated multiplication. For every non-negative integer $n$,

$$(a+b)^n=\sum_{r=0}^{n}\binom nr a^{n-r}b^r.$$

It gives the complete expansion, but its greater strength is that it can find one chosen term or coefficient directly.

Before you start

You should be able to:

• use the laws of indices
• expand and simplify algebraic expressions
• distinguish a term from its coefficient
• evaluate factorials
• substitute negative values with care

Review indices and polynomials if needed. The summation symbol is explained in sigma notation.

Why binomial coefficients appear

Consider

$$(a+b)^3=(a+b)(a+b)(a+b).$$

To produce $a^2b$, choose $b$ from exactly one of the three brackets and $a$ from the other two. There are three possible choices, so its coefficient is $3$. To produce $ab^2$, choose the two brackets supplying $b$. Again there are three choices. Hence

$$(a+b)^3=a^3+3a^2b+3ab^2+b^3.$$

In general, the term containing $b^r$ arises by choosing $b$ from $r$ of the $n$ brackets. The number of ways to do this is

$$\binom nr={}^{n}C_r=\frac{n!}{r!(n-r)!}.$$

Here $0!=1$, so $\binom n0=\binom nn=1$. Also,

$$\binom nr=\binom n{n-r},$$

because choosing the $r$ brackets that supply $b$ is equivalent to choosing the other $n-r$ brackets that supply $a$.

Pascal’s triangle

For small powers, the coefficients can be read from Pascal’s triangle:

$$\begin{array}{c|ccccccc} n&\multicolumn{7}{c}{\text{coefficients}}\\ \hline 0&1\\ 1&1&1\\ 2&1&2&1\\ 3&1&3&3&1\\ 4&1&4&6&4&1\\ 5&1&5&10&10&5&1\\ 6&1&6&15&20&15&6&1 \end{array}$$

Each interior entry is the sum of the two entries above it. Row $n$ contains $n+1$ coefficients, beginning and ending with $1$.

Example 1: use Pascal’s triangle

Expand $(x+2)^5$.

Row $5$ has coefficients $1,5,10,10,5,1$. The power of $x$ decreases from $5$ to $0$, while the power of $2$ increases from $0$ to $5$:

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

Quick check. Substituting $x=0$ gives $2^5=32$, matching the constant term. Substituting $x=1$ makes the coefficients sum to $3^5=243$, and indeed $1+10+40+80+80+32=243$.

The general term

Number the terms from $1$, but number $r$ from $0$. The $(r+1)$th term of $(a+b)^n$ is

$$\boxed{T_{r+1}=\binom nr a^{n-r}b^r},\qquad r=0,1,\ldots,n.$$

This indexing matters. The first term uses $r=0$, the second uses $r=1$, and the $k$th term uses $r=k-1$.

Across consecutive terms:

• the power of $a$ falls by $1$
• the power of $b$ rises by $1$
• the sum of those powers remains $n$

Example 2: expand a binomial with coefficients

Expand $(2x-3)^4$.

Treat the expression as $(2x+(-3))^4$. Keeping $-3$ inside brackets preserves the alternating signs:

$$\begin{aligned} (2x-3)^4 &=\binom40(2x)^4+\binom41(2x)^3(-3) +\binom42(2x)^2(-3)^2\\ &\quad+\binom43(2x)(-3)^3+\binom44(-3)^4\\ &=16x^4-96x^3+216x^2-216x+81. \end{aligned}$$

The sign alternates because $(-3)^r$ is positive for even $r$ and negative for odd $r$.

Example 3: find one specified term

Find the fifth term in the expansion of $(3x-2)^7$.

The fifth term corresponds to $r=4$, not $r=5$:

$$\begin{aligned} T_5 &=\binom74(3x)^{7-4}(-2)^4\\ &=35(27x^3)(16)\\ &=15120x^3. \end{aligned}$$

Only the requested term was calculated.

Finding a coefficient of a chosen power

Write the general term first, then solve an index equation for the required power. This is safer than expanding everything.

Example 4: coefficient of a power of $x$

Find the coefficient of $x^5$ in $(2-3x)^8$.

The general term is

$$T_{r+1}=\binom8r2^{8-r}(-3x)^r =\binom8r2^{8-r}(-3)^r x^r.$$

For an $x^5$ term, $r=5$. Therefore its coefficient is

$$\binom85 2^3(-3)^5 =56\cdot8\cdot(-243) =-108864.$$

The term is $-108864x^5$, while the coefficient is $-108864$.

Example 5: powers arising from both parts

Find the coefficient of $x^4$ in

$$\left(x^2+\frac{2}{x}\right)^5.$$

The general term is

$$\begin{aligned} T_{r+1} &=\binom5r(x^2)^{5-r}\left(\frac2x\right)^r\\ &=\binom5r2^r x^{10-2r-r}\\ &=\binom5r2^r x^{10-3r}. \end{aligned}$$

Set the exponent equal to $4$:

$10-3r=4\quad\Longrightarrow\quad r=2.$

Thus the coefficient is

$\binom522^2=10\cdot4=40.$

The common mistake here is to assume that the power of $x$ is simply $r$. Both parts of the binomial contribute powers of $x$.

Example 6: decide whether a term exists

Does the expansion of

$$\left(x^3+\frac1x\right)^6$$

contain a constant term?

Its general term contains

$$(x^3)^{6-r}(x^{-1})^r=x^{18-4r}.$$

A constant term would require

$18-4r=0\quad\Longrightarrow\quad r=\frac92.$

But $r$ must be an integer from $0$ to $6$. Therefore there is no constant term. Do not round $r$ to a nearby integer.

Products involving binomial expansions

When another factor multiplies the expansion, identify every product that can make the requested power.

Example 7: a coefficient in a product

Find the coefficient of $x^3$ in

$$(1+2x)(1-x)^6.$$

An $x^3$ term can arise in two ways:

$$1\times[\text{the }x^3\text{ term}], \qquad 2x\times[\text{the }x^2\text{ term}].$$

In $(1-x)^6$, the coefficient of $x^r$ is $\binom6r(-1)^r$. Hence the required coefficient is

$$\binom63(-1)^3+2\binom62(-1)^2 =-20+30 =10.$$

Taking only one contribution would miss part of the coefficient.

Binomial approximation

A finite binomial expansion can produce a useful numerical approximation. Write the number as $a+b$, where $a^n$ is easy to calculate and $b$ is small. Early correction terms may then give the required accuracy.

Example 8: approximate a power

Use a binomial expansion to approximate $1.01^5$.

Here $1.01=1+0.01$, so

$$\begin{aligned} 1.01^5 &=(1+0.01)^5\\ &=1+5(0.01)+10(0.01)^2+10(0.01)^3 +5(0.01)^4+(0.01)^5\\ &=1.0510100501. \end{aligned}$$

Using only terms up to $(0.01)^2$ gives $1.051$. The omitted terms total $0.0000100501$, so this shortened value is correct to three decimal places.

This use of the theorem still has an exact finite expansion. Infinite expansions for non-integer powers require an additional convergence condition and are covered in the binomial expansion for rational powers.

Common misconceptions

”The powers in each term are the same”

In $\binom nr a^{n-r}b^r$, one power decreases while the other increases. Their sum is always $n$.

”The fourth term uses $r=4$”

The first term uses $r=0$, so the fourth term uses $r=3$. Translate $T_k$ into $r=k-1$ before substituting.

”A negative sign can be added afterwards”

Use $(-3)^r$, not $-3^r$. Brackets determine whether the sign alternates correctly.

”Every requested power must occur”

The exponent equation may give a non-integer $r$ or an integer outside $0\leq r\leq n$. In either case, that term does not exist.

”The coefficient includes the variable”

In $-12x^4$, the term is $-12x^4$ and the coefficient of $x^4$ is $-12$.

Self-check

1. Expand $(x-3)^4$.
2. Find the fourth term of $(2+5x)^6$.
3. Find the coefficient of $x^6$ in $(1-2x^2)^7$.
4. Find the constant term in $\left(x^2+\frac3x\right)^6$.
5. Find the coefficient of $x^4$ in $(1+x)(1-2x)^5$.
6. Explain why $(a+b)^n$ has $n+1$ terms before like terms are combined.

Answers

1. Using coefficients $1,4,6,4,1$,

   $$(x-3)^4=x^4-12x^3+54x^2-108x+81.$$

2. The fourth term uses $r=3$:

   $$T_4=\binom63 2^3(5x)^3=20000x^3.$$

3. Since $(x^2)^r=x^{2r}$, set $2r=6$, giving $r=3$. The coefficient is

   $$\binom73(-2)^3=35(-8)=-280.$$

4. The power of $x$ in the general term is

   $2(6-r)-r=12-3r.$

   Setting this to zero gives $r=4$, so the constant term is

   $$\binom64 3^4=15\cdot81=1215.$$

5. Combine the $x^4$ term from $(1-2x)^5$ with $x$ times its $x^3$ term:

   $$\binom54(-2)^4+\binom53(-2)^3=80-80=0.$$

   The two contributions cancel, so there is no $x^4$ term after simplification.

6. The index $r$ takes the $n+1$ values $0,1,\ldots,n$, producing one term for each possible number of choices of $b$.

Exam-ready summary

For a positive integer $n$,

$$(a+b)^n=\sum_{r=0}^{n}\binom nr a^{n-r}b^r, \qquad \binom nr=\frac{n!}{r!(n-r)!}.$$

To find a particular term or coefficient:

1. write $T_{r+1}=\binom nr a^{n-r}b^r$
2. simplify the variable power
3. solve for the integer $r$
4. substitute $r$ and evaluate the numerical coefficient
5. check signs, term numbering and the range $0\leq r\leq n$

Next, study the binomial expansion for rational powers, where the expansion may be infinite and valid only for a stated range of values.