The Binomial Theorem | Moosa Academy

Expanding $(a+b)^5$ by multiplying out five brackets is slow and error-prone. The binomial theorem writes the answer down directly. It has only two moving parts — the coefficients, which come from Pascal's Triangle, and the exponents, which follow a fixed pattern in every single case.

Concept When the theorem applies

The binomial theorem expands any expression of the form $(a+b)^n$ , provided there are exactly two terms inside the bracket and $n$ is a positive whole number. The small cases are worth recognising on sight:

(a+b)^0 = 1

(a+b)^1 = a + b

(a+b)^2 = a^2 + 2ab + b^2

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

(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

(a+b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5

Part 1 The coefficients

The numbers in front of each term are read straight off Pascal's Triangle — or equivalently calculated as combinations. Pick the row matching your value of $n$ :

For

(a+b)^4

, take the row

1,\ 4,\ 6,\ 4,\ 1

. Those five numbers are the coefficients, in order.

Part 2 How the exponents move

The exponents follow the same pattern in every expansion:

The exponent on

a

starts at

n

and decreases:

n,\ n-1,\ n-2,\ \dots

down to 0.

The exponent on

b

starts at 0 and increases up to

n

In every term, the two exponents add up to $n$ .

Take $(a+b)^5$ . Ignoring coefficients, the terms are:

a^5,\quad a^4b,\quad a^3b^2,\quad a^2b^3,\quad ab^4,\quad b^5

The power of $a$ falls, the power of $b$ rises, and each pair sums to 5. That last fact is the quickest way to check your work.

Concept How many terms

The expansion of $(a+b)^n$ always has $n + 1$ terms.

n = 2

gives 3 terms.

n = 3

gives 4 terms.

n = 4

gives 5 terms.

n = 5

gives 6 terms.

This is also how you pick the right row of Pascal's Triangle: choose the row containing exactly $n + 1$ numbers.

Concept The general formula

(a+b)^n = \sum_{k=0}^{n} \binom{n}{k}\, a^{\,n-k}\, b^{\,k}

\binom{n}{k}

is the coefficient — the entry from Pascal's Triangle, or the combination "n choose k".

a^{\,n-k}

carries the falling exponent.

b^{\,k}

carries the rising one.

Their exponents sum to

(n-k) + k = n

, exactly as the pattern requires.

The formula is simply the two patterns written in one line. Reading it that way makes it far easier to remember than treating it as a symbol to memorise.

Example Expanding (a + b)⁵

n = 5

, so there will be 6 terms.

Take row 5 of Pascal's Triangle:

1,\ 5,\ 10,\ 10,\ 5,\ 1

Write the powers of

a

falling from 5 and the powers of

b

rising from 0.

Attach each coefficient to its term in order.

⟹ a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵

Example When the second term is a number

Expand $(x + 3)^4$ .

Here

a = x

and

b = 3

, with

n = 4

Row 4 of the triangle gives

1,\ 4,\ 6,\ 4,\ 1

The powers of 3 must be worked out:

3^1 = 3

3^2 = 9

3^3 = 27

3^4 = 81

Multiply each coefficient by the matching power:

4 \times 3 = 12

6 \times 9 = 54

4 \times 27 = 108

1 \times 81 = 81

⟹ x⁴ + 12x³ + 54x² + 108x + 81

Whenever $b$ is a number rather than a letter, remember to raise it to the power as well — leaving it as a plain 3 is the most common error here.

Note Mistakes to avoid

Using the wrong row of Pascal's Triangle — the row must hold

n + 1

numbers.

Letting both exponents rise, or both fall; one always does the opposite of the other.

Ending up with a term whose exponents do not add to

n

Forgetting to raise a numerical second term to its power.

Writing

n

terms instead of

n + 1

Applying the theorem to a bracket with three terms — it is for binomials only.

Summary

The binomial theorem expands (a + b)ⁿ for a positive whole number n.
The coefficients come from the matching row of Pascal's Triangle, or from combinations.
The exponent on a falls from n to 0 while the exponent on b rises from 0 to n.
In every term the two exponents add up to n — a quick way to check your work.
The expansion always has n + 1 terms, which is also how you choose the right row.