General term of binomial expansion — finding specific terms: Visual walkthrough

We assume you know only three plain ideas, all linked at the bottom: what a factorial is, what $\binom{n}{r}$ counts, and how those counts sit in Pascal's triangle. Everything else we earn on the page.

Step 1 — What does " $(a + b)^{n}$ " even mean as a picture?

WHAT. The little exponent $n$ is just shorthand for "multiply this bracket by itself $n$ times."

$(a + b)^{n} = n brackets (a + b) (a + b) \dots (a + b)$

Read each symbol:

$a$ and $b$ — two things being added inside one bracket (they could be $2 x$ and $3$ , or $x^{2}$ and $\frac{1}{x}$ ; for now just "thing one" and "thing two").
$n$ — how many identical brackets stand in a row.

WHY. A power looks scary, but a row of identical brackets is something we can walk through one at a time. That walk is the whole idea.

PICTURE. Below: three brackets drawn as three boxes (the case $n = 3$ ). Inside each box live the two choices $a$ and $b$ .

Figure — General term of binomial expansion — finding specific terms

Step 2 — How is ONE term born? Walk the brackets and pick.

WHAT. To multiply out, you march through the brackets left to right and, from each bracket, grab exactly one of its two letters — either $a$ or $b$ . String your picks together and multiply. That product is one term of the answer.

WHY. Multiplication of sums works exactly this way: $(a + b) (a + b) = aa + ab + ba + bb$ . Each of those four terms is one path of choices — "grab first, grab second." So every term of $(a + b)^{n}$ is one choosing-path through the row of brackets.

PICTURE. One highlighted path: from box 1 take $a$ , from box 2 take $b$ , from box 3 take $a$ . The path spells $a \cdot b \cdot a = a^{2} b$ .

Step 3 — Only the COUNT of $b$ 's matters, not the order.

WHAT. Suppose along your walk you happened to pick $b$ from exactly $r$ of the brackets. Then the other $n - r$ brackets must have handed you $a$ . Multiply:

$n - r of them a \cdot a \dots a \times r of them b \cdot b \dots b = a^{n - r} b^{r}$

Read each symbol:

$r$ — the number of brackets that gave a $b$ . It can be any whole number from $0$ (no $b$ 's at all) up to $n$ (every bracket gave $b$ ).
$n - r$ — forced on us: whatever wasn't a $b$ was an $a$ .
The exponents always sum: $== (n - r) + r = n ==$ . A term can never carry more or fewer than $n$ letters, because there are exactly $n$ brackets.

WHY. Multiplication doesn't care about order: $b \cdot a \cdot b = a \cdot b \cdot b$ . So the paths " $b$ from box 1 & 3" and " $b$ from box 1 & 2" and " $b$ from box 2 & 3" all spell the identical product $a^{1} b^{2}$ . The letters used depend only on how many $b$ 's you took — the count $r$ — never on which boxes.

PICTURE. Three different paths (three different box-choices) that all collapse to the same product $a b^{2}$ because each picked $b$ exactly twice.

Step 4 — Count the identical paths: that count IS $(r n)$ .

WHAT. For a fixed $r$ , how many different walks spell $a^{n - r} b^{r}$ ? That's the same as asking: out of $n$ brackets, in how many ways can I choose the $r$ that hand me a $b$ ? The answer has a name — the combination $(r n)$ .

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

Read each symbol:

$(r n)$ — "choose $r$ boxes out of $n$ ." A plain counting number, like $(2 3) = 3$ .
$n!$ — the factorial, $n! = n \cdot (n - 1) \dots 2 \cdot 1$ ; it counts orderings.
$r!$ and $(n - r)!$ in the denominator — divide out orderings we don't want to double-count, since order of the $b$ 's doesn't matter.

WHY this tool and not just listing? For $n = 3$ you could list paths by hand. For $n = 20$ there are over a million paths — hopeless. $(r n)$ answers "how many identical copies?" in a single number, which is exactly the multiplier each term needs. These are the numbers living in Pascal's triangle.

PICTURE. For $n = 3, r = 2$ : all $(2 3) = 3$ ways to place two $b$ 's among three boxes, shown as three coloured selections — hence the term $a b^{2}$ appears 3 times.

Step 5 — Sweep $r$ over all values → the whole expansion.

WHAT. Every possible term is captured by some value of $r$ . Let $r$ run over every whole number from $0$ to $n$ and add the families up:

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

Read each symbol:

$r = 0 \sum n$ — "add the following, once for each $r$ from $0$ to $n$ ." The $\sum$ (Greek capital sigma, an S for "Sum") is just a stack of plus-signs written compactly.
The lower label $r = 0$ — start with the all- $a$ term ( $a^{n}$ ).
The upper label $n$ — stop at the all- $b$ term ( $b^{n}$ ).

WHY. Nothing is left out (every term has some number of $b$ 's between $0$ and $n$ ) and nothing is counted twice (each $r$ names a different power-pair). So the sum is the full binomial theorem.

PICTURE. A number line of $r$ from $0$ to $n$ ; above each tick sits its term $(r n) a^{n - r} b^{r}$ , and the left column of a strip of Pascal's triangle lines up with the coefficients.

Step 6 — Name each term: why " $T_{r + 1}$ " and not " $T_{r}$ "?

WHAT. We label the terms $T_{1}, T_{2}, T_{3}, \dots$ in the order they appear. But the counter $r$ started at $0$ . So the term for $r = 0$ is the first term, $T_{1}$ ; the term for $r = 1$ is $T_{2}$ ; in general the term for a given $r$ is the $(r + 1)$ -th:

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

Read the offset:

$r$ — starts at $0$ (the counting index).
$r + 1$ — the human term-number, which starts at $1$ .

WHY. Humans count "1st, 2nd, 3rd…" but the exponent of $b$ counts "0, 1, 2…". The little $+ 1$ is just the bridge between those two habits. Miss it and every term-lookup is off by one.

PICTURE. Two rulers side by side: the top ruler ( $r = 0, 1, 2, \dots$ ) is the "how many b's" ruler; the bottom ruler ( $T_{1}, T_{2}, \dots$ ) is the "which term" ruler, shifted right by one.

Step 7 — The edge cases: check the two ends and a zero.

WHAT & WHY. A formula is only trustworthy if it survives its extremes. Feed the boxed formula its smallest and largest $r$ :

$r = 0$ (took no $b$ 's): $T_{1} = (0 n) a^{n} b^{0} = 1 \cdot a^{n} \cdot 1 = a^{n}$ . ✔ The first term is pure $a^{n}$ — matches the single all- $a$ path.
$r = n$ (took every $b$ ): $T_{n + 1} = (n n) a^{0} b^{n} = 1 \cdot 1 \cdot b^{n} = b^{n}$ . ✔ The last term is pure $b^{n}$ .
Total count of terms: $r$ ranges $0, 1, \dots, n$ — that's $== n + 1 ==$ values, so $== n + 1 ==$ terms.
Degenerate $n = 0$ : $(a + b)^{0} = 1$ ; the formula gives only $r = 0$ , i.e. $T_{1} = (0 0) a^{0} b^{0} = 1$ . ✔ One term, and it's $1$ .

Two little facts used above:

$b^{0} = 1$ and $a^{0} = 1$ — anything (nonzero) to the power $0$ is $1$ , i.e. "no letters chosen from that side."
$(0 n) = (n n) = 1$ — there is exactly one way to choose nothing, and exactly one way to choose everything (the outer edge of Pascal's triangle).

PICTURE. The two extreme paths drawn: the all- $a$ walk (every box gives $a$ , giving $a^{n}$ ) and the all- $b$ walk (every box gives $b$ , giving $b^{n}$ ), sitting at the two ends of the row.

The one-picture summary

Every arrow of the derivation, compressed: brackets → pick $b$ from $r$ of them → one product $a^{n - r} b^{r}$ → counted $(r n)$ times → summed over all $r$ → labelled $T_{r + 1}$ .

Recall Feynman retelling — the whole walk in plain words

You have $n$ boxes in a row, and every box hides a red ball ( $a$ ) and a blue ball ( $b$ ). To build a single term you walk the row and snatch one ball from each box. Multiply what you grabbed — that's one term. Here's the trick: multiplication doesn't care about order, so the only thing that decides your product is how many blue balls you grabbed. Call that number $r$ ; then you automatically grabbed $n - r$ reds, so the product is $a^{n - r} b^{r}$ . But loads of different walks grab the same number of blues — precisely $(r n)$ of them — so that product shows up $(r n)$ times, giving $(r n) a^{n - r} b^{r}$ . Let $r$ sweep from $0$ (no blues, pure $a^{n}$ ) up to $n$ (all blues, pure $b^{n}$ ), add everything, and you've rebuilt $(a + b)^{n}$ with no expanding at all. Finally, since $r$ counts from $0$ but people count terms from $1$ , the term for a given $r$ is the $(r + 1)$ -th, written $T_{r + 1}$ .

Connections

3.3.13 General term of binomial expansion — finding specific terms (Hinglish)
Binomial Theorem for positive integer index
Binomial coefficients and Pascal's Triangle
Combinations nCr
Factorials
Middle term of a binomial expansion
Greatest coefficient and greatest term