The Binomial Theorem Explained Simply

Expanding \((a + b)^2\) is easy, but what about \((a + b)^7\) or finding a single term in \((2x - 3)^{10}\)? The binomial theorem gives you a clean formula that does the work without multiplying brackets over and over. This guide explains the theorem simply, shows where the coefficients come from, and works through the question types you will meet in FSc, A Level and first-year university.

What Is a Binomial?

A binomial is an expression with two terms, such as \(a + b\), \(x - 2\) or \(1 + 3x\). The binomial theorem tells us how to expand \((a + b)^n\).

Pascal's Triangle

The coefficients in binomial expansions form Pascal's triangle. Each number is the sum of the two numbers above it:

n = 0: 1 n = 1: 1 1 n = 2: 1 2 1 n = 3: 1 3 3 1 n = 4: 1 4 6 4 1 n = 5: 1 5 10 10 5 1

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

Pascal's triangle is quick for small powers, but for larger \(n\) we use the formula.

The Binomial Theorem

For a positive integer \(n\):

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

where the binomial coefficient is

$$ \binom{n}{r} = \frac{n!}{r!\,(n - r)!} $$

Written out:

$$ (a + b)^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + b^n $$

Patterns to Notice

• There are \(n + 1\) terms.
• The power of \(a\) falls from \(n\) to 0 while the power of \(b\) rises from 0 to \(n\).
• In every term the powers add up to \(n\).
• The coefficients are symmetric: \(\binom{n}{r} = \binom{n}{n-r}\).

Worked Example 1: A Full Expansion

Expand \((2x + 3)^4\).

Step 1: Here \(a = 2x\), \(b = 3\), \(n = 4\). The coefficients from Pascal's triangle are \(1, 4, 6, 4, 1\).

Step 2: Write each term:

$$ (2x)^4 + 4(2x)^3(3) + 6(2x)^2(3)^2 + 4(2x)(3)^3 + 3^4 $$

Step 3: Simplify each term:

• \((2x)^4 = 16x^4\)
• \(4 \times 8x^3 \times 3 = 96x^3\)
• \(6 \times 4x^2 \times 9 = 216x^2\)
• \(4 \times 2x \times 27 = 216x\)
• \(3^4 = 81\)

Answer:

$$ (2x + 3)^4 = 16x^4 + 96x^3 + 216x^2 + 216x + 81 $$

Check: Put \(x = 1\). Left side \(5^4 = 625\). Right side \(16 + 96 + 216 + 216 + 81 = 625\). Correct.

The General Term

The \((r + 1)\)th term of \((a + b)^n\) is

$$ T_{r+1} = \binom{n}{r} a^{n-r} b^r $$

This lets you find any single term without writing the full expansion.

Worked Example 2: Finding a Specific Coefficient

Find the coefficient of \(x^3\) in \((1 - 2x)^8\).

Step 1: Here \(a = 1\), \(b = -2x\), \(n = 8\). The general term is

$$ T_{r+1} = \binom{8}{r}(1)^{8-r}(-2x)^r = \binom{8}{r}(-2)^r x^r $$

Step 2: For \(x^3\), set \(r = 3\).

Step 3: Calculate:

$$ \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56, \qquad (-2)^3 = -8 $$

$$ \text{Coefficient} = 56 \times (-8) = -448 $$

Worked Example 3: The Term Independent of \(x\)

Find the term independent of \(x\) in \(\left(x^2 + \dfrac{1}{x}\right)^6\).

Step 1: The general term is

$$ T_{r+1} = \binom{6}{r}(x^2)^{6-r}\left(\frac{1}{x}\right)^r = \binom{6}{r} x^{12 - 2r} x^{-r} = \binom{6}{r} x^{12 - 3r} $$

Step 2: For the constant term, the power of \(x\) must be zero:

$$ 12 - 3r = 0 \quad \Rightarrow \quad r = 4 $$

Step 3: Calculate:

$$ T_5 = \binom{6}{4} = \frac{6!}{4!\,2!} = 15 $$

The term independent of \(x\) is 15.

Negative and Fractional Powers

For any rational \(n\) and \(|x| < 1\), the expansion becomes an infinite series:

$$ (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots $$

Worked Example 4

Expand \(\sqrt{1 + x}\) up to the \(x^2\) term and use it to estimate \(\sqrt{1.02}\).

Step 1: Here \(n = \frac{1}{2}\):

$$ (1 + x)^{\frac{1}{2}} = 1 + \frac{1}{2}x + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}x^2 + \cdots = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \cdots $$

Step 2: Put \(x = 0.02\):

$$ 1 + 0.01 - \frac{1}{8}(0.0004) = 1.01 - 0.00005 = 1.00995 $$

The true value is \(1.009950\ldots\), so the approximation is excellent. This expansion is valid only for \(|x| < 1\).

Common Mistakes

• Dropping the coefficient inside the bracket. In \((2x)^3\), cube the 2 as well: \(8x^3\).
• Losing negative signs. Keep \(b = -2x\) as one unit with its sign.
• Confusing \(r\) with the term number. \(T_{r+1}\) is the \((r + 1)\)th term, so the 4th term uses \(r = 3\).
• Ignoring the validity condition for negative or fractional powers.

Key Takeaways

• \((a + b)^n\) has \(n + 1\) terms with coefficients \(\binom{n}{r}\), found from Pascal's triangle or the formula.
• The general term \(T_{r+1} = \binom{n}{r}a^{n-r}b^r\) finds any specific term or coefficient.
• For a constant term, set the total power of \(x\) to zero and solve for \(r\).
• For negative or fractional \(n\), the expansion is infinite and valid only when \(|x| < 1\).

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