$(a+b)^n$: binomial theorem and Pascal's triangle

The formula

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

Each term: a binomial coefficient $\binom{n}{k}$, a decreasing power of $a$, an increasing power of $b$.

Pascal’s triangle (up to $n = 5$)

$$\begin{array}{c} 1 \\ 1 \; 1 \\ 1 \; 2 \; 1 \\ 1 \; 3 \; 3 \; 1 \\ 1 \; 4 \; 6 \; 4 \; 1 \\ 1 \; 5 \; 10 \; 10 \; 5 \; 1 \end{array}$$

Each number = sum of the two above. Row $n$ gives the $\binom{n}{k}$ for $k = 0, \dots, n$.

Example: $(x + 2)^4$

Row 4 coefficients: $1, 4, 6, 4, 1$.

$(x + 2)^4 = x^4 + 4 \cdot x^3 \cdot 2 + 6 \cdot x^2 \cdot 4 + 4 \cdot x \cdot 8 + 16$

$= x^4 + 8x^3 + 24x^2 + 32x + 16$

Well-known special cases

• $n = 2$: $(a+b)^2 = a^2 + 2ab + b^2$
• $n = 3$: $(a+b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3$

The final check

Verify your result with $a = b = 1$: sum of coefficients should be $2^n$. For $n = 4$: $1 + 4 + 6 + 4 + 1 = 16 = 2^4$ ✓.

The sign trick

For $(a - b)^n$, alternate signs: $+, -, +, -, \dots$.

$(x - 1)^3 = x^3 - 3x^2 + 3x - 1$.

?Your turn

What is the coefficient of x²y³ in (x + y)⁵?

A.5B.10C.15D.20

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