Answers to exercises

Exercise 1
\(2,\ 5,\ 26,\ 677,\ 458\,330\).
Exercise 2
  1. \(a_n = 2n\)
  2. \(a_n = n^2\).
Exercise 3

The sequence simplifies to

\[ \log_5 2,\ \ 2\log_5 2,\ \ 3\log_5 2,\ \dots \]

and so the general term is \(a_n = n\log_5 2\).

Exercise 4

Here \(a = 210\) and \(d = -13\), so the general term is given by \(a_n = 210 - 13(n-1) = 223-13n\). The equation \(223 - 13n = 12\) has solution \(n = \dfrac{211}{13}\), which is not positive integer. Hence 12 is not a term in the sequence.

Exercise 5

We have \(a = \sqrt{6}\) and \(r = \dfrac{2\sqrt{3}}{\sqrt{6}} = \sqrt{2}\). Thus \(a_n = \sqrt{6}(\sqrt{2})^{n-1} = \sqrt{3}(\sqrt{2})^{n}\).

Exercise 6
\begin{align*} \sum_{k=1}^n \Bigl(\dfrac{1}{k} - \dfrac{1}{k+1}\Bigr) &= \Bigl(1 - \dfrac{1}{2}\Bigr) + \Bigl(\dfrac{1}{2} - \dfrac{1}{3}\Bigr) + \Bigl(\dfrac{1}{3} - \dfrac{1}{4}\Bigr) + \dots + \Bigl(\dfrac{1}{n} - \dfrac{1}{n+1}\Bigr) \\ &= 1 - \dfrac{1}{n+1} = \dfrac{n}{n+1}. \end{align*}
Exercise 7

Writing the series forwards and backwards, we have

\[ \begin{align} S_n = & a & + & (a+d) & + & (a+2d) & + & \cdots & + & (\ell-d) & + & \ell \\ S_n = & \ell & + & (\ell-d) & + & (\ell-2d) & + & \cdots & + & (a+d) & + & a. \end{align} \]

Adding in pairs gives

\[ 2S_n = (a+\ell) + (a+\ell) + \dots + (a+\ell) = n(a+\ell). \]

Hence, \(S_n = \dfrac{n}{2}(a+\ell)\).

Exercise 8

Here \(a = \log_2 3\) and \(d = \log_2 3\), so

\begin{align*} S_n &= \dfrac{n}{2}\bigl(2\log_2 3 + (n-1)\log_2 3\bigr) = \dfrac{1}{2}n(n+1)\log_2 3. \end{align*}
Exercise 9

We have

\begin{align*} S_n &= a + ar + ar^2 + \cdots + ar^{n-1} \\ r S_n &= ar + ar^2 + \cdots + ar^{n-1} + ar^n. \end{align*}

Subtracting gives

\begin{align*} rS_n - S_n = ar^n - a \quad&\implies\quad S_n(r-1) = a(r^n-1) \\ &\implies\quad S_n = \dfrac{a(r^n-1)}{r-1}, \quad\text{provided } r\neq 1. \end{align*}
Exercise 10

We have \(a = \sqrt{3}\) and \(r = \dfrac{6}{\sqrt{3}} = 2\sqrt{3}\). Hence, \(S_n = \dfrac{\sqrt{3}\bigl((2\sqrt{3})^n - 1\bigr)}{2\sqrt{3} - 1}\).

Exercise 11
  1. The triangles \(AXY\) and \(YXB\) are similar. (They have equal angles, as \(\angle AYB = 90^\circ\).) Thus \[ \dfrac{a}{XY} = \dfrac{XY}{b} \quad\implies\quad XY = \sqrt{ab}. \] Hence, the length \(XY\) is the geometric mean of \(a\) and \(b\).
  2. Let \(C\) be the midpoint of \(AB\). Let \(D\) be the point where the perpendicular to \(AB\) at \(C\) cuts the semicircle. Then \(CD\) is the radius of the semicircle, and so \(CD = \dfrac{1}{2}(a+b)\). Clearly, \(CD \geq XY\), and therefore \(\dfrac{1}{2}(a+b) \geq \sqrt{ab}\).
Exercise 12

The common ratio is

\[ r = \dfrac{1}{1+\sqrt{2}} = \sqrt{2} - 1, \]

and so \(-1 < r < 1\). Hence, the geometric series has a limiting sum, given by

\[ S_\infty = \dfrac{1}{1 - (\sqrt{2} - 1)} = \dfrac{1}{2 - \sqrt{2}} = 1 + \dfrac{\sqrt{2}}{2}. \]
Exercise 13

We can write the decimal \(0.\overline{12}\) as

\[ 0.\overline{12} = 0.12121212\ldots = \dfrac{12}{10^2} + \dfrac{12}{10^4} + \dfrac{12}{10^6} + \dotsb, \]

which is a geometric series with \(a = \dfrac{12}{10^2}\) and \(r = \dfrac{1}{10^2}\). The limiting sum is

\[ 0.\overline{12} = \Bigl( \dfrac{12}{10^2} \Bigr) \times \Bigl( \dfrac{1}{1 - \dfrac{1}{10^2}} \Bigr) = \dfrac{4}{33}. \]
Exercise 14

The geometric mean of \(a_1,a_2,\dots,a_n\) is \(G = \sqrt[n]{a_1a_2\dots a_n}\). So

\begin{align*} \log_b G &= \log_b\bigl((a_1a_2\dots a_n)^{\frac{1}{n}}\bigr) \\ &= \dfrac{1}{n} \log_b(a_1a_2\dots a_n) \\ &= \dfrac{1}{n} \bigl(\log_b a_1 + \log_b a_2 + \dots + \log_b a_n\bigr), \end{align*}

which is the arithmetic mean of \(\log_b a_1,\ \log_b a_2,\ \dots,\ \log_b a_n\).

Exercise 15
  1. The arithmetic mean is \(AM = 4\) and the geometric mean is \(GM = \sqrt[3]{60} \approx 3.91\). The harmonic mean \(HM\) satisfies \[ \dfrac{1}{HM} = \dfrac{1}{3}\Bigl(\dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5}\Bigr) = \dfrac{47}{180} \quad\implies\quad HM = \dfrac{180}{47} \approx 3.83. \] So in this case \(HM \leq GM \leq AM\).
  2. Using the AM–GM inequality, we have \(a+b \geq 2\sqrt{ab}\), and so \[ HM = \dfrac{2ab}{a+b} \leq \dfrac{2ab}{2\sqrt{ab}} = \sqrt{ab} = GM. \]
Exercise 16
  1. Let \(f(x) = \log_e x - x\), for \(x > 0\). Then \(f'(x) = \dfrac{1}{x} - 1\), so the only stationary point is at \(x = 1\). We could use a sign diagram to check that \(f(x)\) has a maximum at \(x = 1\). The maximum is \(f(1) = -1\). So, for all \(x > 0\), we have \(\log_e x - x \leq -1\), giving \(\log_e x \leq x - 1\).
  2. Define \(A = \dfrac{1}{n}(a_1 + a_2 + \dots + a_n)\). Substituting \(x = \dfrac{a_1}{A},\ x = \dfrac{a_2}{A},\ \dots,\ x = \dfrac{a_n}{A}\) into the inequality \(\log_e x \leq x - 1\) from part (a) gives \begin{align*} \log_e\Bigl(\dfrac{a_1}{A}\Bigr) &\leq \dfrac{a_1}{A} - 1 \\ \log_e\Bigl(\dfrac{a_2}{A}\Bigr) &\leq \dfrac{a_2}{A} - 1 \\ &\ \vdots\\ \log_e\Bigl(\dfrac{a_n}{A}\Bigr) &\leq \dfrac{a_n}{A} - 1. \end{align*} Adding, we have \[ \log_e\Bigl(\dfrac{a_1}{A}\Bigr) + \log_e\Bigl(\dfrac{a_2}{A}\Bigr) + \dots + \log_e\Bigl(\dfrac{a_n}{A}\Bigr) \leq \dfrac{1}{A}\bigl(a_1 + a_2 + \dots + a_n\bigr) - n. \] Hence, \[ \log_e\Bigl(\dfrac{a_1a_2\dots a_n}{A^n}\Bigr) \leq n-n = 0. \]
  3. Exponentiating both sides of the inequality above gives \[ \dfrac{a_1a_2\dots a_n}{A^n} \leq 1 \quad\implies\quad a_1a_2\dots a_n \leq A^n, \] from which we have \[ (a_1a_2\dots a_n)^{\dfrac{1}{n}} \leq A = \dfrac{a_1 + a_2 + \dots + a_n}{n}, \] which is the desired result.