1. Pascal's triangle
Start with a triangle of numbers. Row 0 is just $1$. Each subsequent row begins and ends with $1$; every interior entry is the sum of the two entries directly above it.
Now expand a few powers of $(a+b)$ by hand:
$$ \begin{aligned} (a+b)^0 &= 1 \\ (a+b)^1 &= a + b \\ (a+b)^2 &= a^2 + 2ab + b^2 \\ (a+b)^3 &= a^3 + 3a^2b + 3ab^2 + b^3 \\ (a+b)^4 &= a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \end{aligned} $$The coefficients are the rows of the triangle. That's not a coincidence — it's the entire content of the binomial theorem, said visually.
A few patterns worth noticing:
- Symmetry. Each row reads the same forward and backward.
- Row sums. Row $n$ sums to $2^n$: $1, 2, 4, 8, 16, 32, 64, \ldots$
- Powers of $11$. The first few rows are the digits of $11^n$ — $11, 121, 1331, 14641$ — and the pattern breaks once entries reach two digits and carry.
- Hockey-stick, Fibonacci, primes… the triangle is a hive of patterns. The ones above are the ones you'll meet in algebra; many more await.
2. Binomial coefficients
The numbers in the triangle have names and a formula.
The number of ways to choose $k$ objects from a set of $n$ (where order doesn't matter). Spoken "$n$ choose $k$." Equal to the entry in row $n$, position $k$ (counting from $0$) of Pascal's triangle, and given by the formula
$$ \binom{n}{k} = \frac{n!}{k!\,(n-k)!} $$Plug in some values to check against the triangle: $\binom{4}{0} = 1$, $\binom{4}{1} = 4$, $\binom{4}{2} = \tfrac{24}{2 \cdot 2} = 6$, $\binom{4}{3} = 4$, $\binom{4}{4} = 1$. Row 4 of the triangle: $1, 4, 6, 4, 1$. ✓
Three facts to internalize:
$$ \binom{n}{0} = \binom{n}{n} = 1, \qquad \binom{n}{k} = \binom{n}{n-k}, \qquad \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k} $$The first says the edges of the triangle are $1$. The second is its left–right symmetry. The third is the addition rule that builds each row from the one above — Pascal's identity, written formally.
3. The theorem
For any non-negative integer $n$ and any numbers $a, b$,
$$ (a + b)^n \;=\; \sum_{k=0}^{n} \binom{n}{k}\, a^{n-k}\, b^{k} $$Written out: $(a+b)^n = a^n + \binom{n}{1} a^{n-1}b + \binom{n}{2} a^{n-2}b^2 + \cdots + \binom{n}{n-1} a b^{n-1} + b^n$.
Three things happen in lockstep as $k$ goes from $0$ to $n$:
- The power of $a$ goes down: $n, n-1, n-2, \ldots, 0$.
- The power of $b$ goes up: $0, 1, 2, \ldots, n$.
- The coefficient walks across row $n$ of Pascal's triangle.
The exponents always sum to $n$ — that's the easiest way to spot-check an expansion.
To expand $(a+b)^7$, write down row 7 of Pascal's triangle ($1, 7, 21, 35, 35, 21, 7, 1$), then attach $a^7, a^6b, a^5b^2, \ldots, b^7$ to each in turn: $$ (a+b)^7 = a^7 + 7a^6b + 21a^5b^2 + 35a^4b^3 + 35a^3b^4 + 21a^2b^5 + 7ab^6 + b^7 $$ No multiplication, no FOIL chain — just lookup-and-attach.
4. Proof by counting
Why does the formula work? Write $(a+b)^n$ as $n$ copies of the factor $(a+b)$ multiplied together:
$$ (a+b)^n = \underbrace{(a+b)(a+b)\cdots(a+b)}_{n\ \text{factors}} $$To expand, you walk through each factor and pick either the $a$ or the $b$. Multiply your $n$ choices. Then add the products over every possible sequence of choices.
How many sequences give the term $a^{n-k} b^k$? Exactly the number of ways to choose which $k$ of the $n$ factors contribute their $b$ (the rest automatically give $a$). That's $\binom{n}{k}$, by definition.
So the $a^{n-k} b^k$ term shows up $\binom{n}{k}$ times, and
$$ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k}\, a^{n-k} b^k. $$The "coefficient" really is a count — that's the whole insight. Binomial coefficients aren't a labeling convention forced onto algebra; they're literally the number of ways the matching term arises.
5. Proof by induction
The combinatorial proof is the prettier one, but the inductive proof shows how Pascal's identity drives the whole structure.
Base case ($n = 0$). $(a+b)^0 = 1$, and $\sum_{k=0}^{0} \binom{0}{k} a^{-k} b^k = \binom{0}{0} = 1$. ✓
Inductive step. Assume $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$. Multiply both sides by $(a+b)$:
$$ (a+b)^{n+1} = a \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k + b \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k $$Distribute the $a$ and the $b$ into each sum:
$$ = \sum_{k=0}^{n} \binom{n}{k} a^{n+1-k} b^k + \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k+1} $$In the second sum, reindex with $j = k+1$ so it becomes $\sum_{j=1}^{n+1} \binom{n}{j-1} a^{n+1-j} b^j$, then rename $j$ back to $k$. Combine the two sums:
$$ (a+b)^{n+1} = a^{n+1} + \sum_{k=1}^{n} \left[\binom{n}{k} + \binom{n}{k-1}\right] a^{n+1-k} b^k + b^{n+1} $$Pascal's identity says $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$. Substituting:
$$ (a+b)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} a^{n+1-k} b^k. \quad\blacksquare $$The induction works because Pascal's identity is precisely the rule that takes the row-$n$ coefficients to the row-$(n+1)$ coefficients — the same rule the triangle is built by.
6. The general term
Sometimes you don't need the whole expansion — you just want the coefficient of one particular term. The $(k+1)$-th term in $(a+b)^n$ is
$$ T_{k+1} = \binom{n}{k}\, a^{n-k}\, b^k. $$(The off-by-one in the subscript is a convention — the first term is $T_1$, with $k=0$.) To pick out a specific term, work out which $k$ gives the powers you want, then read off the coefficient.
Example. Find the coefficient of $x^3$ in $(1 + x)^{10}$. Here $a = 1$, $b = x$, $n = 10$. The general term is $\binom{10}{k}\, 1^{10-k}\, x^k = \binom{10}{k}\, x^k$. We want $k = 3$, so the coefficient is $\binom{10}{3} = \tfrac{10!}{3!\,7!} = 120$.
7. Where you'll use it
Small-$x$ approximations
$(1 + x)^n \approx 1 + nx$ when $x$ is small, because the higher-order terms $\binom{n}{2} x^2 + \cdots$ are negligible compared to $nx$. This is the start of the Taylor series story and the engineering approximation $(1 + x)^n \approx 1 + nx$ that shows up everywhere from physics to finance.
Probability
If you flip a biased coin $n$ times and each flip is heads with probability $p$, the probability of getting exactly $k$ heads is
$$ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}. $$Those coefficients aren't an accident — they're counting the number of head-tail sequences that produce exactly $k$ heads. The binomial theorem and the binomial distribution share their formula for the same reason.
Identity proofs
The classic identity $\sum_{k=0}^{n} \binom{n}{k} = 2^n$ pops out by plugging $a = b = 1$ into the theorem. The alternating sum $\sum_{k=0}^{n} (-1)^k \binom{n}{k} = 0$ for $n \geq 1$ comes from $a = 1, b = -1$. Many identities about binomial coefficients are just clever substitutions into $(a+b)^n$.
8. Common pitfalls
The sum runs from $k = 0$ to $k = n$, inclusive — that's $n + 1$ terms, not $n$. The first term has $k = 0$ (so $a^n$, no $b$), and the last has $k = n$ (so $b^n$, no $a$). The general term $T_{k+1}$ uses $k+1$, not $k$, in its subscript by tradition.
$\binom{n}{0} = \binom{n}{n} = 1$ for every $n$. Forgetting the $a^n$ and $b^n$ end terms in an expansion is a common slip — the binomial coefficients $1$ get absorbed visually and you lose them.
Writing $(a - b)^n$ as $(a + (-b))^n$ means $b$ becomes $-b$ in every term, so odd powers of $b$ pick up a minus sign. The expansion alternates: $$ (a - b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^k = \sum_{k=0}^{n} (-1)^k \binom{n}{k} a^{n-k} b^k. $$ Sign errors here are by far the most common binomial-theorem mistake.
For $(2x + 3)^5$, the coefficient of $x^2$ is not just $\binom{5}{2}$. The general term is $\binom{5}{k}(2x)^{5-k}(3)^k = \binom{5}{k}\, 2^{5-k}\, 3^k\, x^{5-k}$. For $x^2$ you need $5 - k = 2$ so $k = 3$, giving $\binom{5}{3}\, 2^2\, 3^3 = 10 \cdot 4 \cdot 27 = 1080$. Always include the powers of everything inside the parentheses.
9. Worked examples
Example 1 · Expand $(x + 2)^5$
$a = x$, $b = 2$, $n = 5$. Pascal row 5: $1, 5, 10, 10, 5, 1$.
Attach $x^{5-k} \cdot 2^k$ to each coefficient:
$$ \begin{aligned} (x+2)^5 &= 1 \cdot x^5 \cdot 2^0 + 5 \cdot x^4 \cdot 2^1 + 10 \cdot x^3 \cdot 2^2 + 10 \cdot x^2 \cdot 2^3 + 5 \cdot x \cdot 2^4 + 1 \cdot 2^5 \\ &= x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32 \end{aligned} $$Spot check. Plug $x = 1$: $(1+2)^5 = 243$. Sum of coefficients: $1 + 10 + 40 + 80 + 80 + 32 = 243$. ✓
Example 2 · Coefficient of $x^3$ in $(1 + x)^{10}$
General term: $T_{k+1} = \binom{10}{k}\, x^k$. We want $k = 3$.
$$ \binom{10}{3} = \frac{10!}{3!\,7!} = \frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120 $$The coefficient of $x^3$ is $\boxed{120}$.
Example 3 · Show that $\sum_{k=0}^{n} \binom{n}{k} = 2^n$
Apply the binomial theorem with $a = 1$, $b = 1$:
$$ (1 + 1)^n = \sum_{k=0}^{n} \binom{n}{k} \cdot 1^{n-k} \cdot 1^k = \sum_{k=0}^{n} \binom{n}{k} $$The left side is just $2^n$. So $\sum_{k=0}^{n} \binom{n}{k} = 2^n$. $\;\blacksquare$
Combinatorial reading. The right side counts subsets of an $n$-element set (each element is either in or out — two choices per element, $n$ elements, $2^n$ total). The left side counts subsets by size: $\binom{n}{k}$ subsets of size $k$, summed over all sizes. Same set counted two ways.