Binomial theorem — expansion, general term: Worked examples

The scenario matrix

Every binomial-expansion question is really a combination of a few independent "knobs". Here is the full set of knobs and the values they can take:

Knob (what varies)	Possible values	Which example hits it
Sign of $b$	$b > 0$ (plain) / $b < 0$ (alternating signs)	Ex 1 ( $+$ ), Ex 3 ( $-$ )
Numeric factor inside $a, b$	none (baby case $1 + x$ ) / present ( $2 x$ , $3/ x^{2}$ )	Ex 1 (none), Ex 3 (both)
Goal type	full expansion / specific power $x^{k}$ / constant term / coefficient only	Ex 1, Ex 2, Ex 3, Ex 6
Parity of $n$	$n$ even (one middle term) / $n$ odd (two middle terms)	Ex 4 (even), Ex 5 (odd)
Degenerate exponent	no valid $r$ exists (term absent!)	Ex 7
Word / real-world	probability-flavoured story	Ex 8
Exam twist	two conditions at once (equal terms / ratio)	Ex 9
Non-standard $n$	$n = 0$ (trivial) / $n$ negative or fractional (infinite series)	Ex 10

The rule for reading this table: pick a value from each row and you have described a possible question. The ten examples below were chosen so that every value in every row appears at least once.

Example 1 — baby case, $b > 0$ , full expansion · (Sign $+$ , no numeric factor, full expansion)

Before this example, here is the coefficient tool we lean on — Pascal's Triangle built from scratch:

Figure — Binomial theorem — expansion, general term

Figure (alt text): the first six rows of Pascal's Triangle drawn as a pyramid of numbers on midnight-navy. Each number is the sum of the two numbers diagonally above it (two green arrows point down into each interior entry). Row $0$ is a single $1$ at the top; row $5$ across the bottom reads $15101051$ — highlighted in yellow because Example 1 uses exactly that row. The edges are all $1$ 's (blue), because $(0 n) = (n n) = 1$ .

Worked example Expand

$(1 + x)^{5}$ completely. Forecast: before reading on — how many terms will there be, and what is the coefficient of the middle term? Write your guesses down.

Step 1. Identify $a = 1$ , $b = x$ , $n = 5$ , so $T_{r + 1} = (r 5) 1^{5 - r} x^{r} = (r 5) x^{r}$ . Why this step? When $a = 1$ , the factor $1^{5 - r} = 1$ vanishes — this is the baby case where the coefficient is purely $(r 5)$ , nothing else. It is the only case where "coefficient $= (r n)$ " is fully true.

Step 2. Read the coefficients off row $5$ of the triangle above (highlighted yellow): $1, 5, 10, 10, 5, 1$ . Why this step? Row $n$ of Pascal's Triangle is the list $(0 n), \dots, (n n)$ — no need to compute each factorial when the row is short.

Step 3. Attach increasing powers of $x$ : $(1 + x)^{5} = 1 + 5 x + 10 x^{2} + 10 x^{3} + 5 x^{4} + x^{5} .$

Verify: There are $n + 1 = 6$ terms ✓. Set $x = 1$ : the left side is $2^{5} = 32$ ; the right side is $1 + 5 + 10 + 10 + 5 + 1 = 32$ ✓. (Setting $x = 1$ sums all coefficients — a classic sanity check.)

Example 2 — find the term containing a target power · (power $x^{k}$ , numeric-free)

Example 3 — negative $b$ AND numeric factors AND constant term · (Sign $-$ , factors present, constant term)

This one hits three matrix cells at once — the "boss fight".

Worked example Find the term independent of

$x$ in $(2 x^{2} - \frac{1}{x})^{9}$ . Forecast: will your answer be positive or negative? Jot a guess — the sign is where most people slip.

Step 1. Write $a = 2 x^{2}$ , $b = - \frac{1}{x}$ , $n = 9$ : $T_{r + 1} = (r 9) (2 x^{2})^{9 - r} (- \frac{1}{x})^{r} .$ Why this step? Substitute the whole signed $b$ , including the minus. Treating $b$ as $+ \frac{1}{x}$ would silently drop $(- 1)^{r}$ .

Step 2. Split each factor into (number)(power of $x$ ): $T_{r + 1} = (r 9) 2^{9 - r} (- 1)^{r} x^{2 (9 - r)} x^{- r} = (r 9) 2^{9 - r} (- 1)^{r} x^{18 - 3 r} .$ Why this step? We collect all $x$ 's into one exponent $18 - 3 r$ so we can set it against a target. The numeric part $2^{9 - r} (- 1)^{r}$ rides along.

Step 3. "Independent of $x$ " means the exponent is $0$ : $18 - 3 r = 0 \Rightarrow r = 6$ . Why this step? A constant term is $x^{0}$ ; solving gives the unique bracket-count that erases $x$ .

Step 4. $T_{7} = (6 9) 2^{3} (- 1)^{6} = 84 \cdot 8 \cdot 1 = 672.$ Why this step? $(6 9) = (3 9) = 84$ ; $2^{9 - 6} = 2^{3} = 8$ ; $(- 1)^{6} = + 1$ (even power ⇒ positive).

Verify: exponent check $18 - 3 (6) = 0$ ✓. Sign: $r = 6$ even, so $(- 1)^{6} = + 1$ , answer positive $= 672$ ✓.

Example 4 — middle term, $n$ EVEN · (parity even)

Worked example Find the middle term of

$(x - \frac{2}{x})^{6}$ . Forecast: even $n = 6$ — one middle term or two? Which index?

Step 1. Identify $a = x$ , $b = - \frac{2}{x}$ , $n = 6$ , so the general term is $T_{r + 1} = (r 6) x^{6 - r} (- \frac{2}{x})^{r} .$ Why this step? Always name $a$ , $b$ , $n$ and write the general term first, exactly as in Ex 1–3. Substituting the signed $b = - \frac{2}{x}$ now guarantees the minus is carried automatically.

Step 2. $n = 6$ is even, so there is a single middle term at position $\frac{n}{2} + 1 = 4$ , i.e. $r = 3$ . Why this step? With $n + 1 = 7$ terms, the exact centre is the $4$ th; $T_{r + 1}$ with $r + 1 = 4$ gives $r = 3$ .

Step 3. $T_{4} = (3 6) x^{6 - 3} (- \frac{2}{x})^{3} = (3 6) x^{3} \cdot (- 2)^{3} x^{- 3} .$ Why this step? Keep the signed $- \frac{2}{x}$ intact; $(- 2)^{3}$ will be negative.

Step 4. $T_{4} = 20 \cdot (- 8) \cdot x^{0} = - 160.$ Why this step? $(3 6) = 20$ , $(- 2)^{3} = - 8$ , and $x^{3} x^{- 3} = x^{0} = 1$ .

Verify: $(3 6) = \frac{6 \cdot 5 \cdot 4}{6} = 20$ ✓. Sign: $r = 3$ odd ⇒ $(- 2)^{3} < 0$ ⇒ negative $= - 160$ ✓.

Example 5 — middle term, $n$ ODD (TWO of them) · (parity odd)

Example 6 — coefficient only, factor present · (coefficient-only goal, numeric factor)

Example 7 — degenerate case: the term does NOT exist · (no valid $r$ )

The matrix has a nasty cell most notes skip: what if the equation for $r$ has no whole-number solution? Then the requested term simply isn't in the expansion.

Example 8 — real-world / probability flavour · (word problem)

Binomial coefficients aren't just algebra — they count. This example uses the same $(r n)$ engine that powers the Binomial Probability Distribution.

Worked example A fair coin is tossed

$6$ times. Using the expansion of $(H + T)^{6}$ as a bookkeeping device, how many of the $2^{6}$ equally-likely outcomes have exactly $2$ heads? Forecast: guess the count out of $64$ total outcomes.

Step 1. Expand symbolically: $(H + T)^{6} = \sum_{r = 0}^{6} (r 6) H^{6 - r} T^{r}$ . Why this step? Each toss picks $H$ or $T$ from one bracket — exactly the "switch" picture. The term $H^{6 - r} T^{r}$ collects all sequences with $6 - r$ heads and $r$ tails.

Step 2. "Exactly $2$ heads" means $6 - r = 2$ , so $r = 4$ (four tails). The matching term is $H^{2} T^{4}$ with coefficient $(4 6) = 15$ . Why this step? We must read the count off the coefficient of the correct term, $H^{2} T^{4}$ , whose index is $r = 4$ . That coefficient $(4 6)$ counts which $4$ of the $6$ tosses are tails.

Step 3. Equivalently, count directly which $2$ tosses are heads: $(2 6) = 15$ . Why this step? By the symmetry $(4 6) = (2 6)$ — choosing the $4$ tails is the same as choosing the $2$ heads left over. Either count gives the same $15$ , so "count heads" and "count tails" agree. This is why students may write $(2 6)$ without error.

Step 4. (Bonus) probability $= \frac{15}{2 ^{6}} = \frac{15}{64}$ . Why this step? Each of the $64$ sequences is equally likely, and $15$ are favourable.

Verify: total sequences $\sum_{r} (r 6) = 2^{6} = 64$ ✓ (set $H = T = 1$ ). Favourable $(4 6) = (2 6) = 15$ ✓.

Example 9 — exam twist: two conditions at once · (equal-coefficients twist)

Worked example In

$(1 + x)^{n}$ the coefficients of $x^{r}$ and $x^{r + 1}$ are equal, and this happens for $r = 7$ when $n = 15$ . Verify it, and find the general rule. Forecast: what relation between $n$ and $r$ forces two neighbouring coefficients to match?

Step 1. Coefficient of $x^{r}$ is $(r n)$ ; of $x^{r + 1}$ is $(r + 1 n)$ . Set them equal. Why this step? This is the baby case, so coefficients are pure binomials — the twist is comparing two of them.

Step 2. Use the ratio $\frac{( r + 1 n )}{( r n )} = \frac{n - r}{r + 1}$ . Setting the coefficients equal means this ratio $= 1$ : $\frac{n - r}{r + 1} = 1 \Rightarrow n - r = r + 1 \Rightarrow n = 2 r + 1.$ Why this step? The ratio of consecutive binomial coefficients is the cleanest tool — it turns a factorial equation into a one-line linear one.

Step 3. So equality happens exactly when $n$ is odd and $r = \frac{n - 1}{2}$ : the two middle coefficients are equal. Why this step? $n = 2 r + 1$ is precisely the odd case where two middle terms exist (Ex 5) — they share the same coefficient by symmetry $(r n) = (r + 1 n)$ .

Step 4. Check the given numbers: $n = 15 \Rightarrow r = \frac{15 - 1}{2} = 7$ ✓, and $(7 15) = (8 15) = 6435$ .

Verify: $(7 15) = 6435$ and $(8 15) = 6435$ are equal ✓; $n = 2 (7) + 1 = 15$ ✓.

Example 10 — non-standard $n$ : trivial $n = 0$ and the infinite series · (non-standard index)

The theorem in the parent note assumed $n$ a positive integer. Two edge cases fall outside that: the trivial $n = 0$ , and negative/fractional $n$ where the expansion becomes an infinite series — the gateway to Taylor & Maclaurin Series.

Worked example (a) Expand

$(a + b)^{0}$ . (b) Write the first three terms of $(1 + x)^{- 1}$ and of $(1 + x)^{1/2}$ , valid for $∣ x ∣ < 1$ . Forecast: how many terms does each have — finite or endless?

Step 1 (trivial case). $(a + b)^{0} = 1$ for any $a, b$ (not both zero). Why this step? With $n = 0$ there is exactly $n + 1 = 1$ term: $(0 0) a^{0} b^{0} = 1$ . The sum stops immediately — the smallest possible expansion.

Step 2 (why the finite formula breaks). For a positive integer $n$ , the factor $(n - r)$ eventually hits $0$ at $r = n + 1$ , killing every later term — that is why the sum stops at $n$ . When $n$ is negative or fractional, $(n - r)$ is never $0$ , so no term is ever killed: the series runs forever. Why this step? This pinpoints the exact mechanism: finiteness comes from a factor becoming zero, and non-integer $n$ removes that zero.

Step 3 (generalised coefficient). For any real $n$ the coefficient uses the falling product (defined above), not factorials: $(r n) = \frac{n ( n - 1 ) ( n - 2 ) \dots ( n - r + 1 )}{r !} .$ Why this step? The usual $(r n) = \frac{n !}{r ! ( n - r )!}$ needs $(n - r)!$ , which is meaningless for a fractional $n$ (there is no " $(- \frac{3}{2})!$ "). Replacing the top with the falling product removes $(n - r)!$ entirely, so the coefficient still makes sense — and for whole $n$ it gives the same number as before.

Step 4 (compute $(1 + x)^{- 1}$ , $n = - 1$ ). The falling-product coefficients are $(0 - 1) = 1$ , $(1 - 1) = \frac{- 1}{1} = - 1$ , $(2 - 1) = \frac{( - 1 ) ( - 2 )}{2} = 1$ . Attaching powers of $x$ : $(1 + x)^{- 1} = 1 - x + x^{2} - \dots (∣ x ∣ < 1) .$ Why this step? The powers of $x$ still climb $x^{0}, x^{1}, x^{2}, \dots$ exactly as in the finite case; only the coefficients now come from the falling product.

Step 5 (compute $(1 + x)^{1/2}$ , $n = \frac{1}{2}$ ). Coefficients $(0 1/2) = 1$ , $(1 1/2) = \frac{1/2}{1} = \frac{1}{2}$ , $(2 1/2) = \frac{\frac{1}{2} \cdot ( - \frac{1}{2} )}{2} = - \frac{1}{8}$ . Hence $(1 + x)^{1/2} = 1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \dots (∣ x ∣ < 1) .$ Why this step? Same recipe — falling product on top, $r!$ below — now applied to $n = \frac{1}{2}$ .

Step 6 (why $∣ x ∣ < 1$ ?). The series has infinitely many terms, so it only equals the function when those terms shrink to nothing. The powers $x^{r}$ shrink only if $∣ x ∣ < 1$ (e.g. $0. 5^{r} \to 0$ ); if $∣ x ∣ \geq 1$ the terms do not die out and the endless sum has no finite value. So the expansion is trustworthy only inside $∣ x ∣ < 1$ — that is the price of infinitely many terms, a price the positive-integer case never pays because its sum is finite. Why this step? A finite sum is always valid; an infinite one needs its terms to vanish. Stating the region where that happens is part of the answer, not an afterthought.

Verify: $(1 + x)^{- 1} = \frac{1}{1 + x}$ is the familiar geometric series $1 - x + x^{2} - \dots$ (valid for $∣ x ∣ < 1$ ) ✓. Squaring $1 + \frac{1}{2} x - \frac{1}{8} x^{2}$ gives $1 + x + O (x^{3})$ , matching $(1 + x)^{1/2}$ squared $= 1 + x$ ✓.

A picture of the whole matrix

Figure (alt text): a graph plotting the exponent of $x$ (vertical axis) against the counter $r$ (horizontal axis) for three of our expansions. The blue line $18 - 3 r$ is $(2 x^{2} \pm \frac{1}{x})^{9}$ (Ex 2/3); the green line $24 - 4 r$ is $(x^{3} + \frac{1}{x})^{8}$ (Ex 7). A dashed yellow horizontal line marks the "constant term" target ( $0$ ) and a dashed red line marks the target $x^{5}$ . Reading a specific term is just picking a dot; "find the $x^{k}$ term" is finding where a coloured line crosses a horizontal target; "no such term" (Ex 7) is when the target height sits between two dots and never lands on one — the red line never touches a green dot.

Active recall

Recall Forecast then verify

Coefficient of $x^{5}$ in $(3 + 2 x)^{7}$ ? ::: $6048$ .
Does $(x^{3} + 1/ x)^{8}$ have an $x^{5}$ term? ::: No — $r = 4.75$ is not an integer.
Term independent of $x$ in $(2 x^{2} - 1/ x)^{9}$ ? ::: $672$ (at $r = 6$ ).
Number of $6$ -toss outcomes with exactly $2$ heads? ::: $(2 6) = (4 6) = 15$ .
When are consecutive coefficients of $(1 + x)^{n}$ equal? ::: When $n = 2 r + 1$ (odd $n$ , the two middle terms).
First three terms of $(1 + x)^{- 1}$ for $∣ x ∣ < 1$ ? ::: $1 - x + x^{2} - \dots$ .

Connections

Binomial theorem — expansion, general term — the parent formula every example uses.
Combinations nCr — why the coefficient counts (Ex 8).
Factorials — how each $(r n)$ is computed (and why they fail for fractional $n$ , Ex 10).
Pascal's Triangle — fastest route to small coefficient rows (Ex 1).
Binomial Probability Distribution — the coin-toss example is one term of it.
Taylor & Maclaurin Series — the infinite series of Ex 10 for non-integer $n$ .
Permutations vs Combinations — order-free choosing behind $(r n)$ .