The Maths Tailor

To find the coefficient of a single power, do not expand. Write the general term $(k n) (first)^{n - k} (second)^{k}$ , collect all the powers of the variable into one exponent that is linear in $k$ , set that exponent equal to the target power, and solve the single resulting equation for $k$ . Evaluate that one term only.

Trigger: 'find the coefficient of $x^{r}$ ' (or 'the constant term') in a power of a two-term bracket, even after an outside factor or a divisibility twist.

Instances:
(i) for a bracket with a negative or reciprocal power, the exponent of $x$ is an arithmetic expression in $k$ (such as $72 - 8 k$ or $6 - 2 k$ ); set it to the target and read off $k$ ;
(ii) if the bracket is multiplied by a power of $x$ first, subtract that shift from the target power before solving for $k$ ;
(iii) for a divisibility or sign question, keep the term in the symbolic form $(k n) a^{n - k} b^{k}$ and reason on its factors rather than its value.

The non-zero constant

k

is chosen so that the coefficients of

x^{6}

in the expansions of

(1 + k x^{2})^{7}

and

(k + x)^{10}

are equal. What is the value of

k

?

(A)

\frac{1}{6}

(B)

6

(C)

\frac{6}{6}

(D)

6

(E)

\frac{30}{30}

(F)

30

For

(1 + k x^{2})^{7}

, the

x^{6}

term comes from

(k x^{2})^{3}

(3 7) (k x^{2})^{3} = 35 k^{3} x^{6} .

For

(k + x)^{10}

, the

x^{6}

term comes from

k^{4} x^{6}

(4 10) k^{4} x^{6} = 210 k^{4} x^{6} .

Setting the coefficients equal:

35 k^{3} = 210 k^{4}

. Since

k \neq = 0

, divide both sides by

k^{3}

k = \frac{35}{210} = \frac{1}{6} .

Hence the answer is A.

Find the value of the constant term in the expansion of

(x^{6} - \frac{1}{x ^{2}})^{12}

(A)

- 495

(B)

- 220

(C)

- 66

(D)

66

(E)

220

(F)

495

The general term in the binomial expansion is

(k 12) (x^{6})^{12 - k} (- \frac{1}{x ^{2}})^{k} = (k 12) (- 1)^{k} x^{6 (12 - k) - 2 k} = (k 12) (- 1)^{k} x^{72 - 8 k}

For the constant term, set the exponent to zero:

72 - 8 k = 0 \Rightarrow k = 9

. The constant term is

(9 12) (- 1)^{9} = - (3 12) = - \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = - 220

Hence the answer is B.

Find the coefficient of the

x^{4}

term in the expansion of

x^{2} (2 x + \frac{1}{x})^{6}

(A) 15
(B) 30
(C) 60
(D) 120
(E) 240

We need the term in

x^{4}

from

x^{2} \cdot (2 x + \frac{1}{x})^{6}

, so we need the

x^{2}

term from

(2 x + \frac{1}{x})^{6}

.

Using the binomial theorem, the general term is

(k 6) (2 x)^{6 - k} (\frac{1}{x})^{k} = (k 6) 2^{6 - k} x^{6 - k} \cdot x^{- k} = (k 6) 2^{6 - k} x^{6 - 2 k}

.

For the power

x^{2}

6 - 2 k = 2 \Rightarrow k = 2

.

The term is

(2 6) \cdot 2^{4} \cdot x^{2} = 15 \cdot 16 \cdot x^{2} = 240 x^{2}

.

Multiplying by

x^{2}

gives

240 x^{4}

, so the coefficient is

240

. The correct answer is E.

Find the coefficient of

x^{2} y^{4}

in the expansion of

(1 + x + y^{2})^{7}

.

(A)

6

(B)

10

(C)

21

(D)

35

(E)

105

(F)

210

Write

(1 + x + y^{2})^{7} = (x + (1 + y^{2}))^{7}

. By the binomial theorem, the

x^{2}

term is

(2 7) x^{2} (1 + y^{2})^{5} = 21 x^{2} (1 + y^{2})^{5}

. We now need the

y^{4}

term in

(1 + y^{2})^{5}

: by the binomial theorem, this is

(2 5) (y^{2})^{2} = 10 y^{4}

. Therefore the

x^{2} y^{4}

term is

21 \times 10 = 210

. Hence the answer is F.

In the simplified expansion of

(2 + 3 x)^{12}

, how many of the terms have a coefficient that is divisible by 12?

(A)

0

(B)

2

(C)

5

(D)

10

(E)

11

(F)

12

(G)

13

The coefficient of

x^{k}

in the expansion of

(2 + 3 x)^{12}

(k 12) 2^{12 - k} 3^{k}

for

0 \leq k \leq 12

.

A number is divisible by

12 = 4 \times 3

if and only if it is divisible by both 4 and 3.

Divisibility by 3: The factor

3^{k}

ensures divisibility by 3 for all

k \geq 1

. For

k = 0

, the coefficient is

(0 12) 2^{12} = 4096

, which is not divisible by 3.

Divisibility by 4: The factor

2^{12 - k}

ensures divisibility by

4 = 2^{2}

for all

k \leq 10

. For

k = 11

, the coefficient is

(11 12) 2^{1} 3^{11} = 12 \cdot 2 \cdot 3^{11}

, which contains

(11 12) = 12 = 4 \times 3

, so it is divisible by 4. For

k = 12

, the coefficient is

(12 12) 3^{12} = 3^{12}

, which is odd, hence not divisible by 4.

Checking the boundary cases:
-

k = 0

: divisible by 4, not by 3. Not divisible by 12.
-

k = 11

12 \cdot 2 \cdot 3^{11}

, divisible by both 4 and 3. Divisible by 12.
-

k = 12

3^{12}

, not divisible by 4. Not divisible by 12.

For

1 \leq k \leq 10

: divisible by 3 (since

k \geq 1

) and by 4 (since

k \leq 10

). All 10 terms are divisible by 12. Together with

k = 11

, there are

10 + 1 = 11

terms. Hence the answer is E.