Binomial theorem — statement, proof by induction: Visual walkthrough

We use only two starting ideas:

Multiplying two sums means "every term of the first, times every term of the second, all added up".
A bracket $(a + b)$ is just "you may pick $a$ or you may pick $b$ ".

Everything else we build.

Step 1 — One bracket: the atom

WHAT. We start at the smallest possible case, $n = 1$ : $(a + b)^{1} = a + b .$

WHY. Every big expansion is built by multiplying more copies of this same atom. If we don't understand one bracket, we can't understand $n$ of them. So we anchor here.

PICTURE. In the figure, one bracket is a fork in the road: the top branch is the choice " $a$ ", the bottom branch is the choice " $b$ ". Two branches → two terms. Nothing is squared, nothing is chosen twice — there is exactly one letter picked.

Figure — Binomial theorem — statement, proof by induction

Term-by-term of $a + b$ :

$a$ — the outcome "I picked $a$ from the one bracket."
$b$ — the outcome "I picked $b$ from the one bracket."

Number of terms $= 2 = n + 1$ with $n = 1$ . Hold that " $n + 1$ " thought.

Step 2 — Two brackets: choices multiply

WHAT. Multiply a second bracket in: $(a + b)^{2} = (a + b) (a + b) .$ Using "every term times every term": $= a \cdot a + a \cdot b + b \cdot a + b \cdot b = aa + ab + ba + bb .$

WHY. This is the first place the counting idea appears. We are not simplifying yet — we keep $ab$ and $ba$ as different paths, because we want to count paths before we collect them. That distinction is the whole secret.

PICTURE. The one fork of Step 1 becomes a tree with 4 leaves. From bracket 1 you branch $a$ / $b$ ; from bracket 2 you branch again. Every complete path from root to leaf is one product:

$path a, a aa path a, b ab path b, a ba path b, b bb$

$2$ brackets $\Rightarrow 2 \times 2 = 4$ paths (leaves). In general $n$ brackets give $2^{n}$ paths.
Each path is a string of $a$ 's and $b$ 's of length $n$ (here length $2$ ).

This is the raw material. Now we collect.

Step 3 — Collecting like terms: why counting appears

WHAT. Group the four paths by how many $b$ 's each contains: $aa (0 b ’s), ab, ba (1 b each), bb (2 b ’s) .$ Collecting: $(a + b)^{2} = a^{2} + 2 ab + b^{2} .$

WHY. The coefficient in front of each grouped term is literally a count of paths that share the same tally of $b$ 's. So the question "what is the coefficient?" becomes the question "how many paths have exactly $k$ letters equal to $b$ ?". That is a counting question — enter combinations.

PICTURE. Sort the 4 leaves into buckets labelled " $0$ b's", " $1$ b", " $2$ b's". Bucket sizes are $1, 2, 1$ — and the bucket size is the coefficient.

Term-by-term of $a^{2} + 2 ab + b^{2}$ :

$a^{2}$ — coefficient $1$ : only one path ( $aa$ ) has zero $b$ 's.
$2 ab$ — coefficient $2$ : two paths ( $ab$ , $ba$ ) have exactly one $b$ .
$b^{2}$ — coefficient $1$ : only one path ( $bb$ ) has two $b$ 's.

Notice $1 + 2 + 1 = 4 = 2^{2}$ : the buckets partition all the paths.

Step 4 — Naming the count: $(k n)$

WHAT. The coefficient of the grouped term $a^{n - k} b^{k}$ is $(k n)$ .

WHY. Every path with exactly $k$ $b$ 's is uniquely described by which brackets the $b$ 's came from — a choice of $k$ items out of $n$ . Counting those choices is the definition of $(k n)$ . No other tool is needed; combinations were invented for exactly this "choose which ones" question.

PICTURE. Redraw the Step-3 buckets but label each bucket $(0 2), (1 2), (2 2)$ and show that choosing "which bracket gives $b$ " is the same as circling positions in a length- $n$ slot strip.

So for general $n$ : $(a + b)^{n} = \sum_{k = 0}^{n} (k n) a^{n - k} b^{k} .$

$(k n)$ — how many paths have exactly $k$ b's (the count).
$a^{n - k}$ — the $n - k$ brackets that handed you an $a$ (power falls).
$b^{k}$ — the $k$ brackets that handed you a $b$ (power rises).
$\sum_{k = 0}^{n}$ — sweep $k$ across all buckets, from "no b's" to "all b's".

The counting argument is done. Exams still want the induction proof, so we now build the same result a second way — by growing $n$ by one bracket at a time — and watch Pascal's rule fall out.

Step 5 — Growth by one bracket: multiply, then split

WHAT. Suppose we already believe the answer for $n$ (call it $P (n)$ ). Multiply by one more bracket: $(a + b)^{n + 1} = (a + b) known: P (n) k = 0 \sum n (k n) a^{n - k} b^{k} .$ Distribute the extra bracket into its two choices — $\times a$ and $\times b$ : $= chose a from new bracket k = 0 \sum n (k n) a^{n - k + 1} b^{k} + chose b from new bracket k = 0 \sum n (k n) a^{n - k} b^{k + 1} .$

WHY. The new bracket offers exactly the two choices of Step 1: take its $a$ (raise the $a$ -power of every old term) or take its $b$ (raise the $b$ -power). Splitting into these two sums keeps the two possibilities separate so we can line them up.

PICTURE. Two shelves of terms. Top shelf = "new bracket gave $a$ " (every $a$ -exponent bumped up by 1). Bottom shelf = "new bracket gave $b$ " (every $b$ -exponent bumped up by 1). The shelves are shifted relative to each other by one power of $b$ .

$a^{n - k + 1} b^{k}$ — top shelf: old term with one more $a$ .
$a^{n - k} b^{k + 1}$ — bottom shelf: old term with one more $b$ (this is the shift).

Step 6 — Re-indexing: making the shelves line up

WHAT. In the bottom-shelf sum, let $j = k + 1$ (so $k = j - 1$ ). Then $b^{k + 1} = b^{j}$ and $a^{n - k} = a^{n - (j - 1)} = a^{n - j + 1} = a^{(n + 1) - j}$ : $\sum_{k = 0}^{n} (k n) a^{n - k} b^{k + 1} = \sum_{j = 1}^{n + 1} (j - 1 n) a^{(n + 1) - j} b^{j} .$ Rename $j \to k$ . Now both shelves carry the identical power $a^{(n + 1) - k} b^{k}$ .

WHY. We can only add two terms if they are like terms — same power of $a$ , same power of $b$ . Re-indexing slides the bottom shelf sideways so its columns sit exactly under the top shelf's columns.

PICTURE. The shifted bottom shelf of Step 5, slid one notch left, so matching powers of $a^{(n + 1) - k} b^{k}$ stand in the same column. Overlap region $1 \leq k \leq n$ ; two lonely ends stick out.

Top shelf runs $k : 0 \to n$ → its lonely end is $k = 0$ : $(0 n) a^{n + 1} = a^{n + 1}$ .
Bottom shelf runs $k : 1 \to n + 1$ → its lonely end is $k = n + 1$ : $(n n) b^{n + 1} = b^{n + 1}$ .
Middle columns $1 \leq k \leq n$ have one term from each shelf to be added.

Step 7 — Pascal's rule: the two columns merge

WHAT. Add the two shelves column by column: $(a + b)^{n + 1} = k = 0 a^{n + 1} + \sum_{k = 1}^{n} = (k n + 1) by Pascal [(k n) + (k - 1 n)] a^{(n + 1) - k} b^{k} + k = n + 1 b^{n + 1} .$ Since $a^{n + 1} = (0 n + 1) a^{n + 1}$ and $b^{n + 1} = (n + 1 n + 1) b^{n + 1}$ , the two lonely ends fold into the sum: $(a + b)^{n + 1} = \sum_{k = 0}^{n + 1} (k n + 1) a^{(n + 1) - k} b^{k} .$ That is $P (n + 1)$ — the theorem with $n$ replaced by $n + 1$ . Growth verified.

WHY. Pascal's rule is the bridge from level $n$ to level $n + 1$ : it says the count of " $k$ b's out of $n + 1$ brackets" is exactly the sum of the two contributions we already had. This is the same rule that builds Pascal's Triangle — each entry is the sum of the two above it.

PICTURE. The famous triangle: each column-merge of Step 6 is one " $↖ + ↗$ " addition producing the entry below. Row $n$ plus a shifted copy of itself makes row $n + 1$ .

$(k n)$ (upper-right parent) + $(k - 1 n)$ (upper-left parent) $= (k n + 1)$ (child).
Rows are the coefficient lists $1; 11; 121; 1331; \dots$

Base case $n = 1$ (Step 1) $+$ this growth step $\Rightarrow$ induction proves it for all $n \geq 1$ .

Step 8 — Edge & degenerate cases (never leave a gap)

WHAT / WHY / PICTURE, all at once, because each is a quick "check the ends":

The picture stacks four mini-cases (the two ends, $b = 0$ , and the alternating-sign row) so you can see nothing was skipped.

The one-picture summary

Everything on this page in a single frame: the doubling tree on the left feeds the bucket-counts $(k n)$ in the middle, which are exactly Pascal's triangle on the right — and reading any row gives the coefficients of $(a + b)^{n}$ .

Recall Feynman retelling — the whole walkthrough in plain words

Line up $n$ light-switches; each switch shows either $a$ or $b$ . Flip them all and you get one word like $abba$ . There are $2^{n}$ possible words. Sort the words into buckets by how many switches show $b$ . The bucket for "exactly $k$ b's" holds $(k n)$ words, and every word in it equals $a^{n - k} b^{k}$ . Add up the buckets and you've expanded $(a + b)^{n}$ . To prove it grows correctly, add one more switch: it either shows $a$ (raise the $a$ -power of every old word) or $b$ (raise the $b$ -power). Lining up matching words, the count for " $k$ b's out of $n + 1$ " is " $k$ b's before, new switch $a$ " plus " $k - 1$ b's before, new switch $b$ " — that's $(k n) + (k - 1 n) = (k n + 1)$ , Pascal's rule. Start at one switch, grow forever: the theorem holds for all $n$ .

Connections

Pascal's Triangle — Steps 3–7 are the triangle being built.
Combinations — nCr — the bucket count $(k n)$ of Step 4.
Mathematical Induction — Steps 5–7 are the growth step; Step 1 is the base.
Factorials — the $n!$ behind $(k n)$ .
General term & middle term of a binomial expansion — reading one bucket $T_{k + 1}$ .
Binomial series for any index — what happens when $n$ isn't a whole number of brackets.