Combinatorics & Binomial

4 methods · 11 questions

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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.

Linked questions (3)

TMUA 2018 Paper 1 · Q7

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 $30$

Worked solution

Answer:A

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.

TMUA 2018 Paper 2 · Q2

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 $495$

Worked solution

Answer:B

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.

TMUA 2019 Paper 2 · Q1

Find the coefficient of the

x^{4}

term in the expansion of

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

A15
B30
C60
D120
E240

Worked solution

Answer:E

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.