Binomial theorem — expansion, general term: Question bank

The formula under attack throughout — where $n$ is a non-negative integer ( $n = 0, 1, 2, \dots$ ) unless a line says otherwise: $(a + b)^{n} = \sum_{r = 0}^{n} (r n) a^{n - r} b^{r}, T_{r + 1} = (r n) a^{n - r} b^{r} .$

Recall (see Combinations nCr): $(r n)$ is the number of ways to choose $r$ brackets to donate $b$ — keep that picture in mind; most traps below are dissolved by returning to it.

True or false — justify

The coefficient of a term always equals

(r n)

False — that holds only when

a

and

b

carry no numeric or power factors (like

(1 + x)^{n}

). In general the coefficient of the term equals the binomial coefficient

(r n)

times the numeric parts of

a^{n - r}

and

b^{r}

(a + b)^{n}

and

(b + a)^{n}

have the same expansion

True as a whole, but term-by-term they run in reverse order: swapping

a

and

b

maps

r \to n - r

, and

(r n) = (n - r n)

keeps the collection identical.

(a - b)^{n}

has all positive coefficients

False — writing

b \to - b

gives

(- 1)^{r}

per term, so the coefficients of the terms alternate:

+, -, +, -, \dots

r

increases.

(a + b)^{n}

always has exactly

n + 1

terms after collecting

True for non-negative integer

n

; each value

r = 0, 1, \dots, n

gives one distinct power pairing, and none collapse together.

The powers of both

a

and

b

increase across the expansion

False — the exponents must sum to

n

, so as

b

's power

r

rises,

a

's power

n - r

must fall. Only one can grow.

(r n) = (n - r n)

, so the row of binomial coefficients is symmetric

True — choosing which

r

brackets give

b

is the same as choosing which

n - r

give

a

; the binomial-coefficient list reads the same forwards and backwards.

(x + 2)^{4}

the middle binomial coefficient is

(2 4) = 6

True as stated — the binomial coefficient is

6

. But the term's actual coefficient is

6 \cdot 2^{2} = 24

, because

b = 2

contributes

2^{r}

. The trap is confusing the two words.

(1 + x)^{n}

evaluated at

x = 1

gives the sum of all term coefficients

True — substituting

x = 1

collapses every power to

1

, leaving

\sum (r n) = 2^{n}

, which is precisely the coefficient sum.

The term

T_{r + 1}

is the

r

-th term of the expansion

False — it is the

(r + 1)

-th term

T_{r + 1}

. The counter

r

starts at

0

, so it always sits one behind the term's position number.

Spot the error

Student writes the general term of

(2 x - \frac{3}{x ^{2}})^{6}

(r 6) (2 x)^{6 - r} (\frac{3}{x ^{2}})^{r}

Error: they dropped the minus sign. The whole signed quantity

- \frac{3}{x ^{2}}

b

, so it must be

(- 3)^{r}

, which alternates the term's sign with

r

Student expands

(x + 2)^{4}

x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1

Error: they used only the binomial coefficients

(r 4)

and forgot

b = 2

. Including the numeric factor

2^{r}

gives

x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16

Student says

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

for the "

r

-th term" and reads

r

off directly

Error: this line is a single term

T_{r + 1}

, not the sum, and its position is

r + 1

. Reading position as

r

shifts every answer by one.

Student solves for the

x^{6}

term in

(x^{2} + \frac{1}{x})^{9}

by setting

18 - 2 r = 6

Error: they forgot the

x^{- r}

from

(\frac{1}{x})^{r}

. The true exponent is

18 - 2 r - r = 18 - 3 r

; setting that to

6

gives

r = 4

Student claims

(3 7) = \frac{7 !}{3 !}

Error: the denominator is missing

(n - r)!

. Correctly

(3 7) = \frac{7 !}{3 ! 4 !} = 35

; see Factorials.

Student writes the middle term of

(1 + x)^{7}

T_{4}

only

Error:

n = 7

is odd, so there are two middle terms,

T_{4}

and

T_{5}

(from

r = 3

and

r = 4

). A single middle term exists only when

n

is even.

Student says the coefficient of

x^{3}

(1 + 2 x)^{5}

(3 5) = 10

Error:

b = 2 x

contributes

2^{3}

, so the term's coefficient is

(3 5) \cdot 2^{3} = 10 \cdot 8 = 80

, not

10

. The binomial coefficient is

10

; the coefficient is

80

Why questions

Why does the counting number

(r n)

appear at all, rather than some other formula

Because each raw product

a^{n - r} b^{r}

arises once for every choice of which

r

brackets supply

b

(see the box figure above); the number of such choices is

(r n)

, so identical products add up to that count. See Permutations vs Combinations for why order is ignored.

Why is the binomial coefficient a combination

(r n)

and not a permutation

Because the

r

chosen brackets are indistinguishable once multiplied — the product

b \cdot b \dots b

doesn't care which order we picked them, so we count sets, not sequences.

Why do we combine the exponents into a single power of

x

before solving a "term containing

x^{k}

" problem

Because only one clean quantity — the total exponent of

x

— can be set equal to a target

k

. Worked minimally: in

(x^{2} + \frac{1}{x})^{9}

a term is

(x^{2})^{9 - r} (x^{- 1})^{r} = x^{18 - 2 r} \cdot x^{- r} = x^{18 - 3 r}

; now one equation

18 - 3 r = k

solves it. Left split as

x^{18 - 2 r} x^{- r}

there is nothing single to set equal to

k

Why does the constant (independent-of-

x

) term come from setting the exponent to

0

Because

x^{0} = 1

; a term with zero net power of

x

carries no

x

at all — e.g.

x^{18 - 3 r}

is constant when

18 - 3 r = 0

, i.e.

r = 6

— which is exactly what "independent of

x

" means.

Why must we keep negative signs inside the power

(- 3)^{r}

rather than factoring them out once

Because the sign flips with each increment of

r

(- 3)^{r}

+

for even

r

and

-

for odd

r

(e.g.

r = 2 \to + 9

r = 3 \to - 27

), so pulling one sign out front would be wrong for half the terms.

Why do the binomial coefficients match a row of Pascal's Triangle

Because Pascal's rule

(r n) = (r - 1 n - 1) + (r n - 1)

mirrors expanding

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

— each term inherits from one bracket picking

a

b

Why is the sum of all binomial coefficients equal to

2^{n}

Because each of the

n

brackets independently chooses

a

b

— two options each,

2^{n}

total selections — and every selection lands in exactly one coefficient. Set

a = b = 1

to see it.

Why does the same

(r n) p^{r} q^{n - r}

shape appear in probability

Because a run of

n

independent trials chooses "success" in some

r

trials exactly

(r n)

ways, matching the bracket-picking count; see Binomial Probability Distribution.

Edge cases

What is

(a + b)^{0}

and does the formula survive it

It equals

1

: the sum has a single term

r = 0

(0 0) a^{0} b^{0} = 1

. The "

n + 1 = 1

term" rule holds, and

0! = 1

keeps

(0 0) = 1

well-defined.

What happens to the expansion when

b = 0

Every term with

r \geq 1

carries

0^{r} = 0

and vanishes, leaving only

r = 0

(0 n) a^{n} = a^{n}

. The theorem correctly reduces to

(a + 0)^{n} = a^{n}

What happens to the expansion when

a = 0

(the mirror case)

By the same reasoning, every term with

n - r \geq 1

carries

0^{n - r} = 0

and vanishes, leaving only

r = n

(n n) b^{n} = b^{n}

. So

(0 + b)^{n} = b^{n}

, exactly the mirror of the

b = 0

case.

Does

(r n)

make sense when

r = 0

r = n

Yes — both equal

1

, since there is exactly one way to choose no brackets or all brackets for

b

. These are the two end coefficients of the row.

Is the general term valid for

r = n

(the last term)

Yes —

T_{n + 1} = (n n) a^{0} b^{n} = b^{n}

, the final term. The formula covers both endpoints, not just the interior.

What if we try

r > n

or negative

r

T_{r + 1}

It's meaningless for finite expansion. We define

(r n) = 0

whenever integer

r > n

r < 0

, because there are literally zero ways to choose more brackets than exist (or a negative number of them). So no such term arises; the counter is strictly

r = 0, 1, \dots, n

Does

(a + b)^{n}

with the same-sign check work for

(1 - 1)^{n}

Yes — setting

a = 1, b = - 1

gives

\sum (r n) (- 1)^{r} = 0

for

n \geq 1

, showing the alternating coefficients cancel exactly. A neat sanity test.

When

n

is odd, how many "middle" terms are there and why

Two — because

n + 1

is even, so no single term sits at the centre; the pair

T_{(n + 1) /2}

and

T_{(n + 3) /2}

straddle the middle.

Can the exponent equation

n - (something) r = k

give a non-integer

r

, and what does that mean

Yes, and it means no such term exists —

r

must be a whole number in

{0, \dots, n}

, so a fractional solution says the requested power of

x

never appears.

Recall One-line self-test before you leave

Name the three places the counter $r$ can trip you up. Answer ::: (1) term position is $r + 1$ not $r$ ; (2) $r$ must be an integer in $[0, n]$ ; (3) exponent of $a$ is $n - r$ , which shrinks as $r$ grows.

Connections

Binomial theorem — expansion, general term — the parent this bank stress-tests.
Combinations nCr — the counting engine behind every "why $(r n)$ " trap.
Permutations vs Combinations — resolves the order-doesn't-matter confusions.
Factorials — the $(r n) = \frac{n !}{r ! ( n - r )!}$ computation traps.
Pascal's Triangle — the symmetry and Pascal-rule "why" questions.
Binomial Probability Distribution — same coefficient shape in a random experiment.