General term of binomial expansion — finding specific terms: Question bank

Recall the anchor you are testing against: $T_{r + 1} = (r n) a^{n - r} b^{r}, r = 0, 1, 2, \dots, n .$ Here $r$ counts ==how many copies of $b$ you picked==, the whole expansion has $== n + 1 ==$ terms, and the powers on $a$ and $b$ always sum to $n$ .

The one picture below shows why each coefficient is a count of choices — keep it in mind as you answer.

Figure — General term of binomial expansion — finding specific terms

True or false — justify

Predict T or F, then say why in one sentence before revealing.

(a + b)^{n}

the term with

r = 0

b^{n}

False. With

r = 0

you pick zero

b

's, so the term is

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

— the first term is all

a

The 8th term of any expansion uses

r = 8

False. Term number is

r + 1

, so the 8th term uses

r = 7

;

r

starts counting from

0

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

is symmetric, so swapping

a

and

b

leaves each term unchanged.

False. Swapping

a \leftrightarrow b

turns

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

into

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

; the whole sum is unchanged but the individual term at a fixed

r

generally changes.

(a + b)^{n}

there are always

n + 1

terms after collecting like powers.

True. As

r

runs

0

n

you get

n + 1

distinct powers

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

, and none of them are like terms, so no collapsing occurs.

For

(a - b)^{n}

the coefficients are the same magnitudes as

(a + b)^{n}

but alternate in sign.

True. Writing

b \to - b

gives

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

, so the sign flips with the parity of

r

"Coefficient of

x^{k}

" and "the term containing

x^{k}

" always mean the same number.

False. The term includes the factor

x^{k}

; the coefficient is just the numerical (and any non-

x

) part multiplying it.

(x + \frac{1}{x})^{n}

with

n

even, the constant term (power of

x

equal to

0

) is guaranteed to exist.

True. Here the net power of

x

n - 2 r

, which equals

0

r = n /2

; when

n

is even that

r

is a valid integer in range, so the term exists.

(x^{2} + x)^{n}

with

n

even, a constant term is still guaranteed to exist.

False. Here the net power of

x

2 (n - r) + r = 2 n - r

, which is

0

only at

r = 2 n

— out of range — so no constant term exists no matter how even

n

is; parity alone never guarantees it.

The middle term of

(a + b)^{10}

T_{6}

True. With

n = 10

even, there is one middle term

T_{n /2 + 1} = T_{6}

, corresponding to

r = 5

(x - \frac{1}{x})^{n}

every term is positive because the two things being subtracted cancel neatly.

False. The factor

(- 1)^{r}

from

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

makes terms alternate in sign; nothing cancels term-by-term.

Spot the error

Each line states a wrong worked step. Say what broke.

"The 4th term of

(3 x + 1)^{6}

has

r = 4

Error: term number

= r + 1

, so the 4th term needs

r + 1 = 4

, i.e.

r = 3

, not

r = 4

"Term of

(2 x + 5)^{4}

T_{r + 1} = (r 4) 2 x^{4 - r} 5^{r}

Error: the whole entry

2 x

is raised, giving

(2 x)^{4 - r} = 2^{4 - r} x^{4 - r}

; the coefficient

2

must carry the exponent too.

"In

(x - \frac{1}{x ^{2}})^{7}

, the general term is

(r 7) x^{7 - r} \frac{1}{x ^{2 r}}

Error: the minus sign was dropped;

(- \frac{1}{x ^{2}})^{r} = (- 1)^{r} x^{- 2 r}

, so a factor

(- 1)^{r}

is missing.

"Powers in

(a + b)^{9}

add to

9

, so the term at

r = 4

a^{4} b^{5}

Error:

a

carries

n - r = 9 - 4 = 5

and

b

carries

r = 4

, giving

a^{5} b^{4}

; the exponents were swapped.

"To find the constant term of

(2 x + 3)^{5}

, set

24 - 3 r = 0

Error: the exponent formula

24 - 3 r

was borrowed from a different bracket; for

(2 x + 3)^{5}

the net power of

x

5 - r

, which is

0

r = 5

, giving the genuine constant term

(5 5) 3^{5} = 243

— so you must recompute the exponent for this bracket rather than reuse someone else's.

"Coefficient of

x^{9}

(x^{2} + 1/ x)^{12}

(5 12) \cdot x^{9}

Error: the coefficient is the number

(5 12) = 792

; the

x^{9}

is the variable part, not part of the coefficient.

"Since

x^{0} = 1

, the term independent of

x

has coefficient

0

Error:

x^{0} = 1

means the power is

0

, not the coefficient; the constant term keeps its full numerical coefficient (e.g.

- 5376

in the parent's Example 3).

(3 9) = (6 9)

, so

T_{4}

and

T_{7}

are equal in

(2 x - 1/ x^{2})^{9}

Error: the binomial coefficients match, but the powers of

x

and the sign factor

(- 1)^{r}

differ, so the full terms are not equal.

Why questions

Why is the term written

T_{r + 1}

rather than

T_{r}

Because

r

starts at

0

: the first term (

T_{1}

) corresponds to

r = 0

, so the term index is always one more than

r

Why does raising a two-term bracket to power

n

produce exactly one

(r n)

per term?

Because

(r n)

counts how many of the

n

brackets you chose

b

from, and all such choices give the identical term

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

, so they add up to one coefficient.

Why must the exponents of

a

and

b

in any single term add to

n

Each of the

n

brackets contributes exactly one factor (

a

b

), so the total number of factors — hence the total power — is always

n

Why do we "collect the power of the variable into one exponent" before solving?

So the whole term's dependence on

x

becomes a single net-power expression (for

(x^{2} + \frac{1}{x})^{12}

it is

24 - 3 r

, where

24

and

3

come from the powers

x^{2}

and

x^{- 1}

in that bracket), which we then set equal to the target exponent

k

and solve for

r

in one step.

Why can a question about a coefficient still be answered without expanding all terms?

Because the general term pinpoints the single

r

producing that power, so we evaluate just

(r n)

and any constants — no full expansion needed.

Why does

(a - b)^{n}

have its signs decided by the parity of

r

Writing

b \to - b

gives

(- b)^{r} = (- 1)^{r} b^{r}

, and

(- 1)^{r}

+ 1

for even

r

and

- 1

for odd

r

— so the sign tracks whether

r

is even or odd.

Why is

r

restricted to integer values

0 \leq r \leq n

r

counts brackets, which is a whole non-negative number no larger than

n

; a non-integer or out-of-range

r

describes no real term.

Edge cases

If solving "net power of

x

= k

" gives

r = 3.5

, what does it mean?

There is no such term:

r

must be a non-negative integer, so a fractional solution means that power of

x

simply does not appear in the expansion.

If the equation for

r

gives

r = - 2

r = n + 1

, what happens?

Out of range means no term exists; only

0 \leq r \leq n

are valid, so that requested power is absent from the expansion.

What is the "general term" of

(a + b)^{0}

Only

r = 0

is allowed, giving the single term

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

; there is exactly

0 + 1 = 1

term.

For odd

n

, how many middle terms are there and at which

r

Two middle terms,

T_{(n + 1) /2}

and

T_{(n + 3) /2}

, coming from

r = \frac{n - 1}{2}

and

r = \frac{n + 1}{2}

Can two different values of

r

give the same power of

x

in a genuine binomial expansion of

(a + b)^{n}

with

a, b

single powers of

x

No — the collected net power is a linear (degree-1) function of

r

, so it is one-to-one; each power comes from at most one

r

(x + \frac{1}{x})^{n}

, when is there a constant term?

The net power is

n - 2 r

, which is

0

only when

n

is even (giving

r = n /2

); for odd

n

there is no term free of

x

What is the coefficient of the very last term of

(a + b)^{n}

, at

r = n

It is

(n n) = 1

, and the term is

b^{n}

(since

a^{n - n} = a^{0} = 1

b = 0

(so we really expand

(a + 0)^{n}

), what does the general term reduce to?

Every

r > 0

term carries

b^{r} = 0

and vanishes; only

r = 0

survives, giving

a^{n}

— consistent with

(a)^{n}

Connections

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