4.1 Binomial expansion for positive integer n; n! and nCr — Practise

Work through these on your own. Before a long expansion, write down the row of coefficients you will use. When a question asks for one term, use the single-term form instead of expanding everything.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Work out $6!$ .
Work out $\binom{9}{2}$ .
Work out $\binom{9}{7}$ without multiplying seven numbers together.
Expand $(1 + x)^6$ .
Expand $(2 + x)^4$ .
Expand $(1 - 3x)^4$ .
Find the coefficient of $x^3$ in the expansion of $(2 + 5x)^6$ .
Find the term in $x^4$ in the expansion of $(3 - 2x)^7$ .
The coefficient of $x^2$ in the expansion of $(1 + kx)^5$ is $160$ . Find the possible values of $k$ .
A gift of £2,000 grows at 6% a year. Use the binomial expansion of $(1.06)^3$ to work out the value after 3 years, to the nearest penny.
Use the binomial expansion of $(1 + 0.02)^5$ to work out $1.02^5$ to 3 decimal places. Show the first four terms of the expansion.
A fair coin is tossed 6 times. The number of orders in which exactly 2 heads can appear is $\binom{6}{2}$ . Work out that number.

Answers and concise solutions

1. The factorial

Answer: $6! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720$ . Each factorial is the number times the factorial before it, so $6! = 6 \times 120$ .

2. The coefficient

Answer: $\binom{9}{2} = \dfrac{9!}{2!\,7!} = \dfrac{9 \times 8}{2} = 36$ . There are $36$ ways to choose $2$ items from $9$ when the order does not matter.

3. The coefficient

Answer: Use the symmetry $\binom{n}{r} = \binom{n}{n-r}$ . Here $\binom{9}{7} = \binom{9}{2} = 36$ . Choosing the $7$ items to keep is the same as choosing the $2$ items to leave behind.

4. The expansion

Answer: Row 6 of Pascal's triangle is $1, 6, 15, 20, 15, 6, 1$ . Attaching the powers of $x$ gives $1 + 6x + 15x^2 + 20x^3 + 15x^4 + 6x^5 + x^6$ . The constant term is $1$ and the outer coefficients are $1$ , as the symmetry requires.

5. The expansion

Answer: Use the coefficients $1, 4, 6, 4, 1$ with $a = 2$ and $b = x$ : $(2 + x)^4 = 16 + 4 \times 8x + 6 \times 4x^2 + 4 \times 2x^3 + x^4 = 16 + 32x + 24x^2 + 8x^3 + x^4$ . The constant term is $2^4 = 16$ , which checks the first term.

6. The expansion

Answer: Here $b = -3x$ , so the signs alternate. With the coefficients $1, 4, 6, 4, 1$ : $(1 - 3x)^4 = 1 - 12x + 54x^2 - 108x^3 + 81x^4$ . The odd powers carry a minus sign because $(-3x)^3 = -27x^3$ .

7. The coefficient

Answer: The term in $x^3$ is $\binom{6}{3}2^{3}(5x)^3 = 20 \times 8 \times 125x^3 = 20{,}000x^3$ . The coefficient is $20{,}000$ . Working out one term takes one line, so there is no need to expand the whole bracket.

8. The term

Answer: The term in $x^4$ is $\binom{7}{4}3^{3}(-2x)^4 = 35 \times 27 \times 16x^4 = 15{,}120x^4$ . The power of $-2x$ is even, so the term is positive. The coefficient is $15{,}120$ .

9. The values of $k$

Answer: The term in $x^2$ is $\binom{5}{2}(kx)^2 = 10k^2x^2$ , so $10k^2 = 160$ . Then $k^2 = 16$ , giving $k = 4$ or $k = -4$ . Both values give the same coefficient, because the term was squared.

10. The value

Answer: Expand $(1 + 0.06)^3 = 1 + 3(0.06) + 3(0.06)^2 + (0.06)^3 = 1 + 0.18 + 0.0108 + 0.000216 = 1.191016$ . Then $2000 \times 1.191016 = 2382.032$ , which is £2382.03 to the nearest penny. The exact value ends in a half-penny amount, so the final rounding is needed.

11. The value to 3 decimal places

Answer: $(1 + 0.02)^5 = 1 + 5(0.02) + 10(0.02)^2 + 10(0.02)^3 + \dots = 1 + 0.1 + 0.004 + 0.00008 + \dots = 1.1040808$ . To 3 decimal places the value is $1.104$ . The fourth term already contributes less than $0.0001$ , so the first three terms give most of the answer.

12. The number of orders

Answer: $\binom{6}{2} = \dfrac{6 \times 5}{2} = 15$ , so there are $15$ orders with exactly 2 heads. This counting is the link to probability: there are $2^6 = 64$ equally likely orders, so the probability of exactly 2 heads is $\frac{15}{64}$ .