Binomial theorem — expansion, general term: Visual walkthrough

Before we start, two symbols must be earned, because the parent note used them freely.

Step 1 — What "multiply brackets" secretly means

WHAT. We write out $(a + b)^{n}$ as $n$ separate brackets standing in a row, one per copy.

$(a + b)^{n} = ↑ n brackets, each offers a choice of a or b (a + b) (a + b) \dots (a + b)$

WHY. Because multiplication of sums follows one rule you already know from $(a + b) (c + d) = a c + a d + b c + b d$ : you take one term from each bracket, in every possible combination, and multiply your picks together. To see the structure, we must first stop collapsing and just look at the row of brackets.

PICTURE. Below, each bracket is a black box. An arrow leaves each box — you are about to reach in and grab either the $a$ or the $b$ from every box.

Figure — Binomial theorem — expansion, general term

The row of $n$ boxes ← the $n$ copies of $(a + b)$ .
Each box has exactly two exits ( $a$ or $b$ ) ← the two terms inside one bracket.

Step 2 — One "pick" = one raw term

WHAT. We make one full choice: reach into every box and grab exactly one letter. Multiply the grabbed letters. That product is a single raw term.

For example, with $n = 4$ , grabbing $a, b, a, b$ gives the raw term $a \cdot b \cdot a \cdot b = a^{2} b^{2} .$

The four factors ← one letter pulled from each of the four boxes.
$a^{2} b^{2}$ ← we just count how many $a$ 's and how many $b$ 's we happened to grab.

WHY. The whole expansion is nothing but the sum of every raw term you can possibly make. So if we understand one raw term, and then count them, we are done. No algebra magic — pure bookkeeping.

PICTURE. One highlighted path through the boxes (in red) is one raw term. Different paths = different raw terms.

Step 3 — Group the raw terms by "how many $b$ 's"

WHAT. We sort all raw terms into piles. Two raw terms go in the same pile when they grabbed $b$ the same number of times. Call that number $r$ .

If a term grabbed $b$ from exactly $r$ boxes, then it grabbed $a$ from the other $n - r$ boxes, so every raw term in that pile equals $a^{n - r} b^{r} .$

$r$ ← the count of boxes that handed us a $b$ (can be $0, 1, 2, \dots, n$ ).
$n - r$ ← "whatever's left", the boxes that handed us an $a$ .
The two exponents always sum to $n$ : $(n - r) + r = n$ . (This is why $a$ 's power shrinks as $b$ 's power grows — there are only $n$ boxes to share.)

WHY. Sorting turns an unmanageable mess of raw products into a small number of piles, one per value of $r$ . Inside a pile every term is identical ( $a^{n - r} b^{r}$ ), so adding a pile just means counting how many terms are in it.

PICTURE. Three example piles for $n = 4$ : the $r = 0$ pile (all $a$ ), an $r = 2$ pile, the $r = 4$ pile (all $b$ ). Every card in a pile shows the same power-product.

Step 4 — Counting a pile is $(r n)$

WHAT. We count how many raw terms land in the pile labelled $r$ . A raw term in that pile is fixed the moment you decide which $r$ of the $n$ boxes gave the $b$ 's. So:

$# (terms in pile r) = (number of ways to choose r boxes out of n) = (r n) .$

$(r n)$ (read " $n$ choose $r$ ") ← the combination count. Order of the chosen boxes does not matter (Step 2 showed the letters commute), which is exactly why we use a combination, not a permutation.
It is computed with factorials as $(r n) = \frac{n !}{r ! ( n - r )!}$ .

WHY THIS TOOL AND NOT ANOTHER. We need "number of ways to pick an unordered set of $r$ boxes." That is the definition of $(r n)$ . A permutation would count $b$ -from-box-1-then-box-3 as different from box-3-then-box-1 — but those give the identical raw term, so we would over-count. Combinations are the exact tool that ignores order.

PICTURE. For $n = 4$ , $r = 2$ : list every choice of 2 boxes out of 4. There are $(2 4) = 6$ of them, drawn as 6 patterns of two red boxes.

Step 5 — Add the piles: the theorem falls out

WHAT. Every raw term lives in exactly one pile, and the piles are labelled $r = 0, 1, 2, \dots, n$ . Summing all piles reconstructs the whole product:

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

$r = 0 \sum n$ ← "add one contribution for each pile", $r$ running over every possible number of $b$ 's.
The pile term ← exactly what Step 4 gave us.

WHY. Nothing is missing (every raw term is in some pile) and nothing is double-counted (each raw term is in one pile). So the sum of piles equals the full expansion. We never memorised a formula — it was forced by counting. ∎

PICTURE. The row of piles for $n = 4$ lined up left to right, each labelled $(r 4) a^{4 - r} b^{r}$ , with a bracket underneath saying " $+$ " — the final assembly.

For $n = 4$ this reads $(a + b)^{4} = (0 4) a^{4} + (1 4) a^{3} b + (2 4) a^{2} b^{2} + (3 4) a b^{3} + (4 4) b^{4} = a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{4} .$

Step 6 — The edge & degenerate cases (never skipped)

WHAT / WHY / PICTURE, all the corners the naive picture must still handle:

PICTURE. The three end/degenerate cases stacked: $r = 0$ card, $r = n$ card, and the lone $n = 0$ card.

Step 7 — Reading off the general term

WHAT. Give the pile labelled $r$ a human-friendly name. Since $r$ starts at $0$ , the pile $r = 0$ is the 1st term, $r = 1$ the 2nd, and so on. Hence pile $r$ is term number $r + 1$ :

$T_{r + 1} = (r n) a^{n - r} b^{r} .$

$T_{r + 1}$ ← the name of the $(r + 1)$ -th term (subscript is one ahead of the counter $r$ ).
Everything else ← the pile from Step 4.

WHY. This is the workhorse for exams: to hunt the term containing $x^{k}$ , write $T_{r + 1}$ as one power of $x$ , set the exponent to $k$ , solve for $r$ . For the constant term, set the exponent to $0$ .

PICTURE. The pile row again, now with a "term counter" ruler underneath: pile $r = 0$ sits over " $T_{1}$ ", pile $r = 1$ over " $T_{2}$ ", … making the "+1 shift" visible.

The one-picture summary

Everything at once: $n$ boxes → choose $b$ in $r$ of them → $(r n)$ ways → each gives $a^{n - r} b^{r}$ → sum over $r$ → the theorem.

Recall Feynman retelling — explain the whole walkthrough to a 12-year-old

Line up $n$ light switches. Each switch is either flipped to " $a$ " or to " $b$ ". Every possible pattern of flips, once you multiply the labels together, is one term of the answer. Two patterns that flip $b$ the same number of times give the identical result $a^{n - r} b^{r}$ — the order didn't matter. So instead of listing patterns, just count how many patterns flip exactly $r$ switches to $b$ : that count is $(r n)$ (" $n$ choose $r$ "). Multiply the count by its shared result, add up over every possible $r$ from $0$ to $n$ , and you have rebuilt $(a + b)^{n}$ — no formula memorised, only careful counting.

Recall Quick self-test

Why is the coefficient of $a^{n - r} b^{r}$ equal to $(r n)$ ? ::: It counts how many of the $n$ brackets we choose $b$ from; all such choices collapse to the same $a^{n - r} b^{r}$ , so we add them. Why $T_{r + 1}$ and not $T_{r}$ ? ::: The pile counter $r$ starts at $0$ , so pile $r = 0$ is the 1st term — the term index runs one ahead. What does $(a + b)^{0}$ give and why? ::: $1$ — zero boxes means one empty choice, $(0 0) a^{0} b^{0} = 1$ .

Connections

Binomial theorem — expansion, general term — the parent result derived here.
Combinations nCr — the counting engine that produced each pile's size.
Permutations vs Combinations — why order-free counting ( $(r n)$ ) is correct.
Factorials — how $(r n)$ is actually computed.
Pascal's Triangle — the pile sizes, row by row.
Binomial Probability Distribution — same $(r n) p^{r} q^{n - r}$ powers a random experiment.
Taylor & Maclaurin Series — extends the idea to $(1 + x)^{α}$ for non-integer $α$ .