Binomial theorem — statement, proof by induction: Worked examples

Why the coefficients look the way they do (a picture first)

Figure — Binomial theorem — statement, proof by induction

How to read it: each dot is a partial choice; the number in it counts the paths reaching it (that is exactly $(k n)$ , the same numbers as a row of Pascal's Triangle). Follow any downward zig-zag: it grabs one letter per bracket. The red highlighted route grabs $b$ three times out of six, landing in the $(3 6) = 20$ pile — visually the fattest column. This is why the coefficient of $a^{n - k} b^{k}$ equals $(k n)$ : it literally counts routes. (And with $n = 0$ there are no steps at all: a single dot, one empty path — that's Ex 0.)

The scenario matrix

Every binomial-theorem question falls into one of these boxes. Each row is a "case class"; the last column names the example that clears it.

#	Case class	What is tricky about it	Example
Z	Degenerate index $n = 0$ $(a + b)^{0} = 1$	zero steps, one term	Ex 0
A	Plain positive terms $(a + b)^{n}$	just turn the crank	Ex 1
B	A minus sign $(a - b)^{n}$	the $(- 1)^{k}$ sign pattern	Ex 2
C	A specific term / power of $x$	solve for $k$ , don't expand all	Ex 3
D	Term independent of $x$ (power $= 0$ )	limiting/degenerate power case	Ex 4
E	Both terms carry $x$ , fractional/negative powers	careful exponent bookkeeping	Ex 5
F	Middle term(s) — even vs odd $n$	two sub-cases $n$ even / odd	Ex 6
G	Evaluating at special points ( $x = \pm 1$ )	sums of coefficients	Ex 7
H	Real-world / numerical approximation	using $(1 + x)^{n}$ for small $x$	Ex 8
I	Exam twist: unknown in the expansion	reverse-engineer $n$ or a constant	Ex 9

We will visit Z→I in order. (In every one, $n \in N = {0, 1, 2, \dots}$ — recall the definition above.)

Case Z — the degenerate index $n = 0$

Worked example Ex 0 (cell Z) — Expand

$(a + b)^{0}$ Forecast: how many terms does $n + 1$ predict when $n = 0$ ? Guess the whole answer.

Count the terms. The expansion has $n + 1 = 0 + 1 = 1$ term — exactly one. Why this step? The "number of terms $= n + 1$ " rule must survive its smallest input; plugging $n = 0$ tests it.
Write the single term. Only $k = 0$ is allowed (the sum $\sum_{k = 0}^{0}$ has one index): $T_{1} = (0 0) a^{0 - 0} b^{0} = 1 \cdot a^{0} \cdot b^{0} = 1$ . Why this step? $(0 0) = 1$ (one way to choose nothing from nothing), and $a^{0} = b^{0} = 1$ by the power rule — even the empty product is $1$ . So there are zero choice-steps and one empty path (the lone dot in the figure).
Answer: $(a + b)^{0} = 1$ for any $a, b$ (this is the standard convention for any base, including the harmless case where $a + b \neq = 0$ ). Why this step? This is the base-of-the-base case: the theorem doesn't break at $n = 0$ , it just returns the constant $1$ , matching ordinary algebra where anything to the power $0$ is $1$ .

Verify: $(0 0) = 1$ , and $\sum_{k = 0}^{0} (k 0) a^{- k} b^{k} = 1$ . Also $(2 + 5)^{0} = 7^{0} = 1$ . ✓

Case A — plain expansion

Worked example Ex 1 (cell A) — Expand

$(2 x + 3)^{3}$ fully Forecast: guess how many terms, and whether the middle coefficients are bigger than the ends.

Write the general term. $T_{k + 1} = (k 3) (2 x)^{3 - k} (3)^{k}$ . Why this step? Everything is one substitution into $T_{k + 1}$ ; here $a = 2 x$ , $b = 3$ , $n = 3$ .
Run $k = 0, 1, 2, 3$ (there are $n + 1 = 4$ terms).
- $k = 0 : (0 3) (2 x)^{3} = 8 x^{3}$
- $k = 1 : (1 3) (2 x)^{2} (3) = 3 \cdot 4 x^{2} \cdot 3 = 36 x^{2}$
- $k = 2 : (2 3) (2 x) (3)^{2} = 3 \cdot 2 x \cdot 9 = 54 x$
- $k = 3 : (3 3) (3)^{3} = 27$ Why this step? Each term multiplies the binomial coefficient by both powers — keep the numbers separate from the $x$ 's.
Assemble: $(2 x + 3)^{3} = 8 x^{3} + 36 x^{2} + 54 x + 27$ . Why this step? Since each $k$ produced a different power of $x$ , no two terms combine — we just line them up in descending powers, so the expansion is simply the four terms added.

Verify: set $x = 1$ : LHS $= (2 + 3)^{3} = 125$ ; RHS $= 8 + 36 + 54 + 27 = 125$ . ✓

Case B — the minus sign

Worked example Ex 2 (cell B) — Expand

$(x - 2)^{4}$ Forecast: which terms come out negative? (Guess before reading.)

Substitute the whole second term including its sign: $b = - 2$ . So $T_{k + 1} = (k 4) x^{4 - k} (- 2)^{k}$ . Why this step? The commonest mistake is to forget the sign. Writing $(- 2)^{k}$ bakes the alternation in automatically: $(- 2)^{k} = (- 1)^{k} 2^{k}$ .
Run $k = 0 \dots 4$ :
- $k = 0 : x^{4}$
- $k = 1 : (1 4) x^{3} (- 2) = - 8 x^{3}$
- $k = 2 : (2 4) x^{2} (4) = 24 x^{2}$
- $k = 3 : (3 4) x (- 8) = - 32 x$
- $k = 4 : (16) = 16$ Why this step? Odd $k$ gives a negative sign (because $(- 1)^{k} = - 1$ ), even $k$ positive — that's the signs alternating.
Assemble: $(x - 2)^{4} = x^{4} - 8 x^{3} + 24 x^{2} - 32 x + 16$ . Why this step? Again each $k$ owns a distinct power of $x$ , so we merely add the five terms; the alternating signs come straight from step 1's $(- 2)^{k}$ .

Verify: set $x = 1$ : LHS $= (1 - 2)^{4} = 1$ ; RHS $= 1 - 8 + 24 - 32 + 16 = 1$ . ✓

Case C — a single specific term

Case D — the term that has NO $x$ (degenerate power = 0)

Worked example Ex 4 (cell D) — Term independent of

$x$ in $(x^{2} - \frac{2}{x})^{6}$ Forecast: "independent of $x$ " means the total power of $x$ equals $0$ . Guess whether such a $k$ even exists.

General term: $T_{k + 1} = (k 6) (x^{2})^{6 - k} (- \frac{2}{x})^{k} = (k 6) (- 2)^{k} x^{2 (6 - k)} x^{- k}$ . Why this step? $\frac{1}{x} = x^{- 1}$ , so the second factor contributes a negative power. We must combine both powers.
Collect the exponent of $x$ : $2 (6 - k) - k = 12 - 3 k$ . Why this step? One symbol ( $x$ ) can only have one power per term; add the exponents.
Set it to zero (that's what "independent of $x$ " means): $12 - 3 k = 0 \Rightarrow k = 4$ . Why this step? If this equation had no whole-number solution in $0 \leq k \leq 6$ , there would be no constant term — a real possibility you must always check. Here $k = 4$ is valid.
Evaluate: $(4 6) (- 2)^{4} = 15 \cdot 16 = 240$ . Why this step? Only $k = 4$ hits power $0$ , so this one term IS the constant term — evaluate its number and you are done.
Answer: $240$ .

Verify: power at $k = 4$ is $12 - 12 = 0$ ✓; value $(4 6) \cdot (- 2)^{4} = 15 \cdot 16 = 240$ ✓.

Case E — both terms carry $x$ , mixed fractional powers

Worked example Ex 5 (cell E) — Coefficient of

$x^{7}$ in $(x + \frac{3}{x ^{2}})^{10}$ Forecast: $x = x^{1/2}$ and $\frac{1}{x ^{2}} = x^{- 2}$ . Guess: will the required $k$ be a whole number?

General term: $T_{k + 1} = (k 10) (x^{1/2})^{10 - k} (3 x^{- 2})^{k} = (k 10) 3^{k} x^{\frac{10 - k}{2}} x^{- 2 k}$ . Why this step? Rewrite every root and reciprocal as a power of $x$ first — never mix notations.
Collect the exponent: $\frac{10 - k}{2} - 2 k = \frac{10 - k - 4 k}{2} = \frac{10 - 5 k}{2}$ . Why this step? Combine into a single power so we can match it to $x^{7}$ .
Match: $\frac{10 - 5 k}{2} = 7 \Rightarrow 10 - 5 k = 14 \Rightarrow k = - \frac{4}{5}$ . Why this step? $k$ must be an integer with $0 \leq k \leq 10$ . Here $k = - 0.8$ is not a valid $k$ .
Conclusion: there is no $x^{7}$ term — its coefficient is $0$ . Why this step? Covering the degenerate outcome: not every power you ask for appears. Always report $0$ when no valid $k$ exists.

Verify: check a power that DOES appear — $k = 2$ gives $\frac{10 - 10}{2} = 0$ , so a constant term exists; $k = 0$ gives $x^{5}$ . The reachable powers are $5, \frac{5}{2}, 0, \dots$ stepping by $- \frac{5}{2}$ ; $x^{7}$ is not in that list. ✓

Case F — the middle term(s): even vs odd $n$

Worked example Ex 6 (cell F) — Middle term(s) of (i)

$(x + y)^{6}$ and (ii) $(x + y)^{7}$ Forecast: an even power gives ONE middle term, an odd power gives TWO. Guess which position(s). (The path picture at the top of this page shows why the middle piles up highest.)

(i) $n = 6$ (even). There are $n + 1 = 7$ terms, so the single middle one is term number $\frac{n}{2} + 1 = 4$ .

Middle term $= T_{4}$ , i.e. $k = 3$ . Why this step? Term index $= k + 1$ , so the $4$ th term uses $k = 3$ — exactly the centre of $0 \dots 6$ .
$T_{4} = (3 6) x^{3} y^{3} = 20 x^{3} y^{3}$ . Why this step? Substituting $k = 3$ into $T_{k + 1}$ with $a = x, b = y$ gives equal powers $x^{3} y^{3}$ — the balanced centre, which the figure shows as the tallest single column. Answer (i): $20 x^{3} y^{3}$ .

(ii) $n = 7$ (odd). There are $8$ terms; the two middle ones are terms $\frac{n + 1}{2} = 4$ and $\frac{n + 3}{2} = 5$ , i.e. $k = 3$ and $k = 4$ . 3. $T_{4} = (3 7) x^{4} y^{3} = 35 x^{4} y^{3}$ , and $T_{5} = (4 7) x^{3} y^{4} = 35 x^{3} y^{4}$ . Why this step? $(3 7) = (4 7) = 35$ — the coefficients are mirror-symmetric about the centre (in the path picture, the two central columns have identical height). Answer (ii): $35 x^{4} y^{3} and 35 x^{3} y^{4}$ .

Verify: (i) $(3 6) = 20$ ✓. (ii) $(3 7) = 35 = (4 7)$ ✓; both middle coefficients equal, as Pascal's Triangle predicts.

Case G — evaluating at special points

Worked example Ex 7 (cell G) — Find

$k = 0 \sum n (k n)$ and $k = 0 \sum n 2^{k} (k n)$ Forecast: these are the same theorem, just fed clever inputs. Guess the first answer ( $2^{n}$ ?).

Start from $(1 + x)^{n} = \sum_{k = 0}^{n} (k n) x^{k}$ (put $a = 1, b = x$ ). Why this step? Sums of binomial coefficients are just the expansion read at a chosen $x$ .
Put $x = 1$ : $(1 + 1)^{n} = 2^{n} = \sum_{k = 0}^{n} (k n)$ . Why this step? At $x = 1$ every $x^{k}$ becomes $1$ , so the expansion collapses to the bare sum of coefficients — instantly $2^{n}$ . So $k = 0 \sum n (k n) = 2^{n}$ — also the number of subsets of an $n$ -element set.
Put $x = 2$ : $(1 + 2)^{n} = 3^{n} = \sum_{k = 0}^{n} (k n) 2^{k}$ . Why this step? Choosing $x = 2$ makes each $x^{k} = 2^{k}$ , exactly the weight we want on $(k n)$ ; the left side evaluates in one shot. So $k = 0 \sum n 2^{k} (k n) = 3^{n}$ .

Verify (n=4): $\sum (k 4) = 1 + 4 + 6 + 4 + 1 = 16 = 2^{4}$ ✓; $\sum 2^{k} (k 4) = 1 + 8 + 24 + 32 + 16 = 81 = 3^{4}$ ✓.

Case H — real-world numerical approximation

Worked example Ex 8 (cell H) — Estimate

$(1.02)^{5}$ using the binomial expansion Forecast: since $0.02$ is tiny, high powers of it are negligible. Guess $(1.02)^{5} \approx 1.1$ ?

Write $1.02 = 1 + x$ with $x = 0.02$ : $(1 + x)^{5} = \sum_{k = 0}^{5} (k 5) x^{k}$ . Why this step? The $(1 + x)^{n}$ form is built for approximations — the terms shrink fast when $x$ is small. This is the finite cousin of the Binomial series for any index (and here $n = 5 \in N$ , so the sum is exact and finite).
Keep only the first three terms (units: pure number, $x$ dimensionless): $(1 + x)^{5} \approx 1 + 5 x + (2 5) x^{2} = 1 + 5 x + 10 x^{2}$ . Why this step? $x^{3} = 8 \times 1 0^{- 6}$ is far below our target precision — dropping it is the whole art of approximation.
Substitute $x = 0.02$ : $1 + 5 (0.02) + 10 (0.0004) = 1 + 0.10 + 0.004 = 1.104$ . Why this step? We evaluate the kept terms and add; the discarded $x^{3}$ and higher would only nudge the 4th decimal, so $1.104$ is safe to 3 places.
Answer: $(1.02)^{5} \approx 1.104$ .

Verify: true value $1.0 2^{5} = 1.1040808 \dots$ ; our estimate $1.104$ is correct to $3$ decimals. ✓

Case I — the exam twist (reverse-engineer)

Worked example Ex 9 (cell I) — Find

$n$ if the coefficients of $x^{2}$ and $x^{3}$ in $(1 + x)^{n}$ are equal Forecast: the coefficient of $x^{k}$ in $(1 + x)^{n}$ is just $(k n)$ . Guess: does equal-coefficients force a specific $n$ ?

Identify the coefficients: coefficient of $x^{2}$ is $(2 n)$ , of $x^{3}$ is $(3 n)$ . Why this step? In $(1 + x)^{n}$ the power of $x$ equals $k$ , so "coefficient of $x^{k}$ " literally means $(k n)$ .
Set them equal: $(2 n) = (3 n)$ . Why this step? Translate the word "equal" into an equation before touching factorials.
Expand and simplify. $\frac{n !}{2 ! ( n - 2 )!} = \frac{n !}{3 ! ( n - 3 )!}$ . Cancel $n!$ and cross-multiply: $3! (n - 3)! = 2! (n - 2)! \Rightarrow 6 (n - 3)! = 2 (n - 2)!$ . Why this step? We clear the fractions so the unknown $n$ is no longer trapped inside a denominator.
Cancel the factorials. Since $(n - 2)! = (n - 2) (n - 3)!$ , divide both sides by $(n - 3)!$ : $6 = 2 (n - 2)$ . Why this step? $(n - 3)!$ is common to both sides, so cancelling it collapses the factorial equation into ordinary algebra — one linear equation in $n$ .
Solve for $n$ . $6 = 2 (n - 2) \Rightarrow 3 = n - 2 \Rightarrow n = 5$ . Why this step? Divide by $2$ , then add $2$ ; $n = 5$ is a valid non-negative integer, so the finite theorem genuinely applies and the answer is admissible.

Verify: in $(1 + x)^{5}$ , $(2 5) = 10$ and $(3 5) = 10$ — equal. ✓

Recall Quick self-test

What is $(a + b)^{0}$ and how many terms does it have? ::: $1$ ; exactly one term ( $n + 1 = 1$ ). Coefficient of $x^{5}$ in $(3 + 2 x)^{8}$ ? ::: $48384$ (from $k = 5$ ). Constant term of $(x^{2} - 2/ x)^{6}$ ? ::: $240$ (from $k = 4$ ). Coefficient of $x^{7}$ in $(x + 3/ x^{2})^{10}$ ? ::: $0$ — no integer $k$ works. $\sum_{k = 0}^{n} 2^{k} (k n) = ?$ ::: $3^{n}$ (put $x = 2$ in $(1 + x)^{n}$ ). Middle term of $(x + y)^{6}$ ? ::: $20 x^{3} y^{3}$ ( $T_{4}$ , $k = 3$ ). Estimate $(1.02)^{5}$ to 3 places? ::: $1.104$ . If $(2 n) = (3 n)$ then $n = ?$ ::: $5$ . Does the finite theorem apply to $(1 + x)^{- 2}$ ? ::: No — negative index gives an infinite series ( $∣ x ∣ < 1$ ).

Connections

General term & middle term of a binomial expansion — the engine behind Cases C–F.
Combinations — nCr and Factorials — every coefficient here.
Pascal's Triangle — the symmetry $(k n) = (n - k n)$ used in Ex 3 and Ex 6, and the path picture at the top.
Binomial series for any index — Case H taken to infinitely many terms for small $x$ ; the tool to use when $n \in / N$ .
Mathematical Induction — why all of this is even guaranteed to hold.