Chapter 4: Review

The chapter problem, solved

The village hall needs a new roof, and it will cost £9,000. Two fundraising streams run at the same time, and a former resident gives £2,000 at the start. The gift is invested at 6% compound interest each year.

Step 1. Name each stream as a sequence. The quiz night raises £150 in month 1 and £50 more each month, so it is arithmetic:

q_n = 150 + 50(n - 1) = 50n + 100.

The online appeal raises £600 in month 1 and 80% of that each month after, so it is geometric:

a_n = 600(0.8)^{n-1}.

Step 2. Find the first few months, to check the models. The quiz night gives $150, 200, 250, \dots$ , and the appeal gives $600, 480, 384, \dots$ . The quiz night grows by a fixed amount, and the appeal falls by a fixed factor. Over a year the quiz night raises more, even though the appeal starts higher.

Step 3. Find the total from each stream after 12 months. For the arithmetic stream,

S^{q}_{12} = \frac{12}{2}\bigl(2(150) + 11(50)\bigr) = 6(300 + 550) = 5100.

For the geometric stream,

S^{a}_{12} = \frac{600(1 - 0.8^{12})}{1 - 0.8} = 3000(1 - 0.8^{12}) = 2793.84.

After one year the fund holds $5100 + 2793.84 = 7893.84$ , which is short of £9,000 by £1106.16.

Step 4. Find the most the appeal can ever raise. The common ratio is $0.8$ , and $|0.8| < 1$ , so the series converges:

S^{a}_\infty = \frac{600}{1 - 0.8} = 3000.

The appeal can never raise more than £3,000, however long it runs. That is why the second stream cannot pay for the roof on its own.

Step 5. Decide whether the fund reaches £9,000, and when. After $n$ months the total is

T_n = 25n^2 + 125n + 3000(1 - 0.8^n).

At the end of month 13,

T_{13} = 5850 + 2835.07 = 8685.07,

which is below £9,000. At the end of month 14,

T_{14} = 6650 + 2868.06 = 9518.06,

which is above. The fund passes £9,000 during month 14. The roof is paid for, and the two streams together do it sooner than the quiz night alone.

Step 6. Refine the plan. The quiz night alone raises $25n^2 + 125n$ pounds. Set this equal to $9000$ :

25n^2 + 125n = 9000 \implies n^2 + 5n - 360 = 0.

The formula gives $n = \frac{-5 + \sqrt{1465}}{2} \approx 16.6$ , so the quiz night alone would take until month 17. The appeal brings the date forward by about three months, and it is worth the effort even though its total is capped at £3,000.

Step 7. Find the value of the gift. After 3 years,

2000(1.06)^3 = 2000(1 + 3(0.06) + 3(0.06)^2 + (0.06)^3) = 2000(1.191016) = 2382.03,

so the gift is worth £2382.03. If the committee moves the money after $2\frac{1}{2}$ years instead, the binomial expansion for a rational power gives

(1.06)^{\frac{5}{2}} \approx 1 + \frac{5}{2}(0.06) + \frac{15}{8}(0.06)^2 + \frac{5}{16}(0.06)^3 = 1.1568175\dots,

so the gift is worth about $2000 \times 1.1568175 = 2313.63$ pounds.

The chapter's answer. The fund passes £9,000 during month 14 and holds £9518.06 by the end of that month. The online appeal raises £2793.84 in the first year and can never exceed £3,000. The quiz night alone would take until month 17. The £2,000 gift is worth £2382.03 after 3 years, or about £2313.63 after $2\frac{1}{2}$ years. The roof is paid for.

Mixed practice — choose the method first

For each question, say which method you will use and why before you calculate. The items mix this chapter with Chapters 1, 2 and 3, so the choice is not always obvious.

Expand $(1 + 2x)^6$ .
Find the coefficient of $x^3$ in the expansion of $(3 - x)^7$ .
Expand $(1 - 3x)^{\frac{1}{2}}$ up to and including the term in $x^3$ , and state the range of validity.
Use the first three terms of the expansion of $(1 + x)^{\frac{1}{2}}$ to estimate $\sqrt{1.04}$ .
Evaluate $\sum_{r=1}^{15}(4r - 3)$ .
Work out $\sum_{r=1}^{12}r^2$ .
Work out the sum of the first 25 terms of $7, 11, 15, \dots$ .
Find the sum to infinity of $18 + 12 + 8 + \dots$ .
A geometric sequence has 2nd term $6$ and 5th term $48$ . Find the first term and the common ratio.
A recurrence is $u_{n+1} = 0.6u_n + 5$ with $u_1 = 10$ . Find the level the terms settle on.
Work out the gradient of the line through $(2, -1)$ and $(6, 7)$ .
Solve $2x^2 + 5x - 3 = 0$ .
Prove that the sum of two odd numbers is even.
A quantity doubles every 3 years. State the model you would use and why.
The appeal model assumes the income falls by exactly 20% each month. State one further assumption and one way the model could fail.

Answers

Binomial theorem, positive whole number $n$ . With coefficients $1, 6, 15, 20, 15, 6, 1$ and $b = 2x$ : $(1 + 2x)^6 = 1 + 12x + 60x^2 + 160x^3 + 240x^4 + 192x^5 + 64x^6$ .
One term at a time. The term in $x^3$ is $\binom{7}{3}3^{4}(-x)^3 = 35 \times 81 \times (-1)x^3 = -2835x^3$ , so the coefficient is $-2835$ .
General binomial series, with $n = \frac{1}{2}$ and $-3x$ in place of $x$ . $(1 - 3x)^{\frac{1}{2}} = 1 - \frac{3x}{2} - \frac{9x^2}{8}
\frac{27x^3}{16} + \dots$, valid for $|x| < \frac{1}{3}$ .
First three terms, then substitute. With $x = 0.04$ , $1 + \frac{1}{2}(0.04) - \frac{1}{8}(0.04)^2 = 1 + 0.02 - 0.0002 = 1.0198$ . So $\sqrt{1.04} \approx 1.0198$ .
Split the sum and use the standard result. $\sum(4r - 3) = 4\sum r - 3n$ with $n = 15$ , so $4(120) - 45 = 480 - 45 = 435$ .
The standard result for squares. $\frac{12 \times 13 \times 25}{6} = \frac{3900}{6} = 650$ .
Arithmetic series. $S_{25} = \frac{25}{2}(2(7) + 24(4)) = 12.5(14 + 96) = 12.5(110) = 1375$ .
Sum to infinity. The ratio is $\frac{12}{18} = \frac{2}{3}$ , and $\left|\frac{2}{3}\right| < 1$ , so $S_\infty = \frac{18}{1 - \frac{2}{3}} = \frac{18}{\frac{1}{3}} = 54$ .
Two terms, then a division. $ar = 6$ and $ar^4 = 48$ , so $r^3 = \frac{48}{6} = 8$ and $r = 2$ . Then $a = \frac{6}{2} = 3$ . The sequence is $3, 6, 12, \dots$ .
Fixed point of the recurrence. At the settling level $L$ , $L = 0.6L + 5$ , so $0.4L = 5$ and $L = 12.5$ . The level exists because $|0.6| < 1$ .
Gradient from two points. $\frac{7 - (-1)}{6 - 2} = \frac{8}{4} = 2$ .
Factorise the quadratic. $(2x - 1)(x + 3) = 0$ , so $x = \frac{1}{2}$ or $x = -3$ .
Deduction with general letters. Let the two odd numbers be $2m + 1$ and $2n + 1$ for whole numbers $m$ and $n$ . Their sum is $2m + 2n + 2 = 2(m + n + 1)$ , which has a factor $2$ . So the sum is even.
A geometric model. Doubling is a fixed factor, not a fixed amount, so $u_n = a(2)^{n-1}$ if the value is measured at the start of each 3-year period. An arithmetic model would add the same amount each period, which does not describe doubling.
State an assumption and its limit. The model assumes every month is the same length and that the appeal income is received on time. It also assumes the fall is exactly 20% each month, which is unlikely in a real appeal: a single mention by a popular account could raise the income instead of lowering it, and the model cannot show that.

What you can now do

use factorials and binomial coefficients, and expand $(a + bx)^n$ for a positive whole number $n$ ;
find a single term or coefficient without expanding the whole bracket;
expand $(1 + x)^n$ for a rational $n$ , state the range of validity, and approximate a value with the first few terms;
generate a sequence from a formula or a recurrence, and describe its behaviour;
read and write sums in sigma notation, and use the standard results;
find the $n$ th term and the sum of an arithmetic sequence;
find the $n$ th term and the sum of a geometric sequence, and the sum to infinity when $|r| < 1$ ;
choose a sequence model, state its assumptions, find when it reaches a target, and refine it.

Your relevance reflection

Return to the prompt in the chapter opener: something in your own life that changes by a rule. Write down your starting value and your rule again. Now add three things you could not have said before this chapter: the formula for the $n$ th value, the total after a given number of steps, and whether the change is a fixed amount or a fixed factor. Finish with one sentence about what the model leaves out.