🔬 Tutorial problems delta

Note

This problems are designed to help you practice the concepts covered in the lectures. Not all problems may be covered in the tutorial, those left are for additional practice.

\(\delta\).1

Formulation

What are the first five terms of each of the following sequences?

(a) \(\left\{ \frac{2n - 1}{3n+2} \right\}_{n \in \mathbb{N}}\)

(b) \(\left\{ \frac{1 - (-1)^n}{n^3} \right\}_{n \in \mathbb{N}}\)

(c) \(\left\{ \frac{(-1)^{n - 1}}{(2)(4)(6)\cdots(2n)} \right\}_{n \in \mathbb{N}}\)

(d) \(\left\{ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{2^n}\right\}_{n \in \mathbb{N}}\)

(e) \(\left\{\frac{(-1)^{n - 1} x^{2n-1}}{(2n-1)!} \right\}_{n \in \mathbb{N}}\)

Solution

(a)

The first five terms of the sequence \(\left\{\frac{2 n-1}{3 n+2}\right\}_{n \in \mathbb{N}}\) are

\[\begin{split} \begin{array}{l} x_{1}=\frac{2(1)-1}{3(1)+2}=\frac{2-1}{3+2}=\frac{1}{5}, \\ x_{2}=\frac{2(2)-1}{3(2)+2}=\frac{4-1}{6+2}=\frac{3}{8}, \\ x_{3}=\frac{2(3)-1}{3(3)+2}=\frac{6-1}{9+2}=\frac{5}{11}, \\ x_{4}=\frac{2(4)-1}{3(4)+2}=\frac{8-1}{12+2}=\frac{7}{14}=\frac{1}{2}, \\ x_{5}=\frac{2(5)-1}{3(5)+2}=\frac{10-1}{15+2}=\frac{9}{17} \text {. } \end{array} \end{split}\]

(b) The first five terms of the sequence \(\left\{\frac{1-(-1)^{n}}{n^{3}}\right\}_{n \in \mathbb{N}}\) are

\[\begin{split} \begin{array}{l} x_{1}=\frac{1-(-1)^{1}}{1^{3}}=\frac{1-(-1)}{1}=\frac{1+1}{1}=\frac{2}{1}=2, \\ x_{2}=\frac{1-(-1)^{2}}{2^{3}}=\frac{1-(1)}{8}=\frac{1-1}{8}=\frac{0}{8}=0, \\ x_{3}=\frac{1-(-1)^{3}}{3^{3}}=\frac{1-(-1)}{27}=\frac{1+1}{27}=\frac{2}{27}, \\ x_{4}=\frac{1-(-1)^{4}}{4^{3}}=\frac{1-(1)}{64}=\frac{1-1}{64}=\frac{0}{64}=0, \\ x_{5}=\frac{1-(-1)^{5}}{5^{3}}=\frac{1-(-1)}{125}=\frac{1+1}{125}=\frac{2}{125} . \end{array} \end{split}\]

(c) The first five terms of the sequence \(\left\{\frac{(-1)^{n-1}}{(2)(4)(6) \cdots(2 n)}\right\}_{n \in \mathbb{N}}\) are

\[\begin{split} \begin{array}{l} x_{1}=\frac{(-1)^{1-1}}{(2)}=\frac{(-1)^{0}}{2}=\frac{1}{2}, \\ x_{2}=\frac{(-1)^{2-1}}{(2)(4)}=\frac{(-1)^{1}}{8}=\frac{-1}{8} \\ x_{3}=\frac{(-1)^{3-1}}{(2)(4)(6)}=\frac{(-1)^{2}}{48}=\frac{1}{(48)}, \\ x_{4}=\frac{(-1)^{4-1}}{(2)(4)(6)(8)}=\frac{(-1)^{3}}{384}=\frac{-1}{384}, \\ x_{5}=\frac{(-1)^{5-1}}{(2)(4)(6)(8)(10)}=\frac{(-1)^{4}}{3,840}=\frac{1}{(3,840)} \end{array} \end{split}\]

(d)

The first five terms of the sequence \(\left\{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{2^{n}}\right\}_{n \in \mathbb{N}}\) are

\[\begin{split} \begin{array}{l} x_{1}=\sum_{i=1}^{1} \frac{1}{2^{i}}=\frac{1}{2^{1}}=\frac{1}{2}, \\ x_{2}=\sum_{i=1}^{2} \frac{1}{2^{i}}=\frac{1}{2^{1}}+\frac{1}{2^{2}}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4} \\ x_{3}=\sum_{i=1}^{3} \frac{1}{2^{i}}=\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=\frac{3}{4}+\frac{1}{8}=\frac{7}{8}, \\ x_{4}=\sum_{i=1}^{4} \frac{1}{2^{i}}=\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\frac{1}{2^{4}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{7}{8}+\frac{1}{16}=\frac{15}{16}, \\ x_{5}=\sum_{i=1}^{5} \frac{1}{2^{i}}=\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\frac{1}{2^{4}}+\frac{1}{2^{5}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=\frac{15}{16}+\frac{1}{32}=\frac{31}{32} . \end{array} \end{split}\]

(e)

The first five terms of the sequence \(\left\{\frac{(-1)^{n-1} x^{2 n-1}}{(2 n-1) !}\right\}_{n \in \mathbb{N}}\) are

\[\begin{split} \begin{array}{l} x_{1}=\frac{(-1)^{1-1} x^{2(1)-1}}{(2(1)-1) !}=\frac{(-1)^{0} x^{2-1}}{((2-1) !}=\frac{(1) x^{1}}{1 !}=\frac{x^{1}}{1}=x, \\ x_{2}=\frac{(-1)^{2-1} x^{2(2)-1}}{(2(2)-1) !}=\frac{(-1)^{1} x^{4-1}}{((4-1) !}=\frac{(-1) x^{3}}{3 !}=\frac{-x^{3}}{(3)(2)(1)}=\frac{-x^{3}}{6}, \\ x_{3}=\frac{(-1)^{3-1} x^{2(3)-1}}{(2(3)-1) !}=\frac{(-1)^{2} x^{6-1}}{((6-1) !}=\frac{(1) x^{5}}{5 !}=\frac{x^{5}}{(5)(4)(3)(2)(1)}=\frac{x^{5}}{120}, \\ x_{4}=\frac{(-1)^{4-1} x^{2(4)-1}}{(2(4)-1) !}=\frac{(-1)^{3} x^{8-1}}{((8-1) !}=\frac{(-1) x^{7}}{7 !}=\frac{-x^{7}}{(7)(6)(5)(4)(3)(2)(1)}=\frac{-x^{7}}{5,040}, \\ x_{5}=\frac{(-1)^{5-1} x^{2(5)-1}}{(2(5)-1) !}=\frac{(-1)^{4} x^{10-1}}{((10-1) !}=\frac{(1) x^{9}}{9 !}=\frac{x^{9}}{(9)(8)(7)(6)(5)(4)(3)(2)(1)}=\frac{x^{9}}{362,880} \end{array} \end{split}\]

\(\delta\).2

Formulation

Use the principle of mathematical induction to prove that

\[ \sum_{i=1}^{n} \frac{1}{i(i+1)}=\frac{n}{(n+1)} \text { for all } n \in \mathbb{N} \]

Hint: Remember that

\[ \sum_{i=1}^{n} \frac{1}{i(i+1)}=\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\cdots+\frac{1}{(n)(n+1)} \]

Solution

Proof: We will use the principle of mathematical induction to prove this claim.

First, consider the case where \(n=1\). Note that

\[ \sum_{i=1}^{n} \frac{1}{i(i+1)}=\sum_{i=1}^{1} \frac{1}{i(i+1)}=\frac{1}{(1)(1+1)}=\frac{1}{(1)(2)}=\frac{1}{2}, \]

and

\[ \frac{n}{(n+1)}=\frac{1}{(1+1)}=\frac{1}{2} \]

Since \(\frac{1}{2}=\frac{1}{2}\), we know that the claim is true when \(n=1\).

Now assume that the claim is true when \(n=k\). This means that

\[ \sum_{i=1}^{k} \frac{1}{i(i+1)}=\frac{k}{(k+1)} \]

Call this “the inductive assumption”.

Next, consider the case when \(n=(k+1)\). Note that

\[\begin{split} \begin{array}{l} \sum_{i=1}^{(k+1)} \frac{1}{i(i+1)} = \left(\sum_{i=1}^{k} \frac{1}{i(i+1)}\right)+\frac{1}{(k+1)((k+1)+1)} \\ = \left(\sum_{i=1}^{k} \frac{1}{i(i+1)}\right)+\frac{1}{(k+1)(k+2)} \\ = \frac{k}{(k+1)}+\frac{1}{(k+1)(k+2)} \text { (from the inductive assumption) } \\ = \frac{k(k+2)}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)} \\ = \frac{k(k+2)+1}{(k+1)(k+2)} \\ = \frac{k^{2}+2 k+1}{(k+1)((k+1)+1)} \\ = \frac{(k + 1)^2}{(k+1)((k+1)+1)} \\ = \frac{(k+1)}{((k+1)+1)} \end{array} \end{split}\]

which is in the desired form. Thus we know that, if the claim is true when \(n=k\), then it is also true when \(n=(k+1)\).

We have shown that if the claim is true when \(n=k\), then it is also true when \(n=(k+1)\). We have also shown that the claim is true when \(n=1\). Thus we can conclude, from the principle of mathematical induction, that the claim must be true for all \(n \in \mathbb{N}\). \(\quad \blacksquare\)

\(\delta\).3

Formulation

Use the principle of mathematical induction to prove that

\[ n^{3}+(n+1)^{3}+(n+2)^{3} \text { is exactly divisible by nine for all } n \in \mathbb{N} \text {. } \]

Hint: A number \(x \in \mathbb{N}\) is exactly divisible by nine if and only if \(\frac{x}{9} \in \mathbb{N}\).

Solution

Proof: We will use the principle of mathematical induction to prove this claim.

First, consider the case where \(n=1\). Note that

\[ 1^{3}+(1+1)^{3}+(1+2)^{3}=1^{3}+2^{3}+3^{3}=1+8+27=36=(9)(4) \text {. } \]

Thus we know that the claim is true when \(n=1\).

Now assume that the claim is true when \(n=k\). This means that

\[ k^{3}+(k+1)^{3}+(k+2)^{3}=9 x \text { for some } x \in \mathbb{N} \text {. } \]

Call this “the inductive assumption”.

Next, consider the case when \(n=(k+1)\). Note that

\[\begin{split} \begin{array}{l} (k+1)^{3}+((k+1)+1)^{3}+((k+1)+2)^{3}\\ = (k+1)^{3}+(k+2)^{3}+(k+3)^{3} \\ = (k+1)^{3}+(k+2)^{3}+(k+3)(k+3)^{2} \\ = (k+1)^{3}+(k+2)^{3}+(k+3)\left(k^{2}+6 k+9\right) \\ = (k+1)^{3}+(k+2)^{3}+\left(k^{3}+6 k^{2}+9 k+3 k^{2}+18 k+27\right) \\ = (k+1)^{3}+(k+2)^{3}+\left(k^{3}+9 k^{2}+27 k+27\right) \\ = k^{3}+(k+1)^{3}+(k+2)^{3}+\left(9 k^{2}+27 k+27\right) \\ = \left(k^{3}+(k+1)^{3}+(k+2)^{3}\right)+9\left(k^{2}+3 k+3\right) \\ = 9 x+9\left(k^{2}+3 k+3\right) (\text { from the inductive assumption }) \\ = 9\left(x+k^{2}+3 k+3\right) . \end{array} \end{split}\]

Since both \(x \in \mathbb{N}\) and \(k \in \mathbb{N}\), we know that \(\left(x+k^{2}+3 k+3\right) \in \mathbb{N}\) as well. Thus we know that, if the claim is true when \(n=k\), then it is also true when \(n=(k+1)\).

We have shown that if the claim is true when \(n=k\), then it is also true when \(n=(k+1)\). We have also shown that the claim is true when \(n=1\). Thus we can conclude, from the principle of mathematical induction, that the claim must be true for all \(n \in \mathbb{N}\). \(\quad \blacksquare\)

\(\delta\).4

Formulation

Consider the geometric progression

\[ \left\{P_{n}\right\}_{n=1}^{\infty}=\left\{a, a r, a r^{2}, \cdots\right\}=\left\{a r^{(n-1)}\right\}_{n=1}^{\infty}, \]

where \(r \notin\{0,1\}\). Note that the sum of the first \(N\) terms of this geometric progression is defined by

\[ S_{N}=\sum_{n=1}^{N} P_{n}=\sum_{n=1}^{N} a r^{(n-1)}=\sum_{k=0}^{N-1} a r^{k} \]

Use the principle of mathematical induction to prove that this partial sum for this arbitrary geometric progression is given by the formula

\[ S_{N}=\frac{a\left(1-r^{N}\right)}{(1-r)} \]

Solution

Proof: Note that the following proof is based on the one that is provided in Basov (2011, pp. 23-24).

First, recall that the sum of the first \(N\) terms of this geometric progression is defined by

\[ S_{N}=\sum_{n=1}^{N} P_{n}=\sum_{n=1}^{N} a r^{(n-1)}=\sum_{k=0}^{N-1} a r^{k} \]

Consider the case when \(N=1\). Note that

\[ S_{1}=\sum_{n=0}^{1-1} P_{n}=\sum_{n=0}^{0} P_{n}=P_{0}=a \]

Note also that

\[\begin{split} \begin{array}{l} \frac{a\left(1-r^{N}\right)}{(1-r)} = \frac{a\left(1-r^{1}\right)}{(1-r)} \\ = \frac{a(1-r)}{(1-r)} \\ = a . \end{array} \end{split}\]

Thus the claim is true when \(N=1\).

Suppose that the claim is true when \(N=k\). This means that

\[ S_{k}=\frac{a\left(1-r^{k}\right)}{(1-r)} \quad\quad (\text { the Inductive Assumption }) . \]

Consider the case in which \(N=(k+1)\). Note that

\[\begin{split} \begin{array}{l} S_{(k+1)} = \sum_{n=0}^{(k+1)-1} P_{n} \\ = \sum_{n=0}^{k} P_{n} \\ = \left(\sum_{n=0}^{k-1} P_{n}\right)+P_{k} \\ = S_{k}+P_{k} . \end{array} \end{split}\]

We know that

\[ S_{k}=\frac{a\left(1-r^{k}\right)}{(1-r)} \]

from our inductive assumption. Furthermore, we know from our arithmetic progression sequence that

\[ P_{k}=a r^{k} \]

Thus we have

\[\begin{split} \begin{array}{l} S_{(k+1)} = S_{k}+P_{k} \\ = \frac{a\left(1-r^{k}\right)}{(1-r)}+a r^{k} \\ = \frac{a\left(1-r^{k}\right)}{(1-r)}+\frac{a r^{k}(1-r)}{(1-r)} \\ = \frac{a\left(1-r^{k}\right)+a r^{k}(1-r)}{(1-r)} \\ = \frac{a-a r^{k}+a r^{k}-a r^{(k+1)}}{(1-r)} \\ = \frac{a-a r^{(k+1)}}{(1-r)} \\ = \frac{a\left(1-r^{(k+1)}\right)}{(1-r)} . \end{array} \end{split}\]

Thus we know that if the claim is true when \(N=k\), then it will also be true when \(N=(k+1)\).

We have shown that the claim is true when \(N=1\). We have also shown that if the claim is true when \(N=k\), then it will also be true when \(N=(k+1)\). Thus we can apply the principle of mathematical induction to conclude that the claim is true for all \(N \in \mathbb{N}\). \(\quad \blacksquare\)

\(\delta\).5

Formulation

The B-Happy corporation runs a lottery with a prize of \(\$ 1,000,000\). The winners are given the following options. They can receive the prize of \(\$ 1,000,000\) immediately, or they can receive a payment of \(\$ 140,000\) every year for the next ten years. The nominal interest rate that is used to discount future receipts is \(i=8 \%\) per annum.

(a) What is the future value of \(\$ 1,000,000\) in ten years time if interest is compounded yearly?

(b) What is the future value of the regular yearly payments of \(\$ 140,000\) made at the end of each of the next ten years if interest is compounded yearly?

(c) Which option should a winner of the lottery choose?

(d) Which option should the winner choose if the payments of \(\$ 140,000\) are made at the start, rather than at the end, of each year?

(e) Explain what happens in parts (a), (b), (c) and (d) of this question if interest is compounded continuously instead of yearly, but payments are still made yearly.

Source: Shannon (1995, pp. 350-351)

Solution

Note: the answers to this question was sourced from the answer key to the exercises in Shannon (1995). This answer key was written by John Shannon and Ted McDonald.

(a)

\[\begin{split} \begin{array}{l} S = P(1 + i)^n \\ = 1000000(1 + 0.08)^{10} \\ = 2158925 \end{array} \end{split}\]

(b)

\[\begin{split} \begin{array}{l} S = R \left[ \frac{(1 + i)^n - 1}{i} \right] \\ = 140000 \left[ \frac{(1 + 0.08)^{10} - 1}{0.08} \right] \\ = 2028118.75 \end{array} \end{split}\]

(c) A winner is better off receiving the prize money of $1000000 immediately as long as all other relevant factors such as the amount of tax paid is the same in both cases.

(d)

\[\begin{split} \begin{array}{l} S = R (1 + i) \left[ \frac{(1 + i)^n - 1}{i} \right] \\ = (2028118.70)(1 + 0.08) \\ = 2190368.20 \end{array} \end{split}\]

The winner is better off accepting the annuity if it is an annuity due with payments at the start of each year.

(e)

When interest is compounded continuously but payments are made yearly we have a general annuity. We use the same formulae we used with ordinary annuities but the interest rate is now the effective interest rate. This is found using the formula for effective interest rates with continuous compounding.

\[ \begin{array}{l} r = e^i - 1 = e^{0.08} - 1 = 0.083287 \text{ or } 8.3287\% \end{array} \]

In part (a)

\[ S = Pe^{rt} = 1000000e^{(0.08)10} = 2225540.90 \]

In part (b)

\[ S = 140000 \left[ \frac{(1 + 0.083287)^{10} - 1}{0.083287} \right] = 2060051.80 \]

As before we chose the immediate payment.

With an annuity due

\[ S = R(1 + r) \left[ \frac{(1 + r)^n - 1}{r} \right] = (2060051.80)(1.083287) = 2231627.30 \]

As before we choose the ‘annuity due’.

\(\delta\).6

Formulation

Consider the following business opportunity. If you pay \(\$ F\) now (in period zero), you can build a production plant that will be ready for operation in the next period. This plant will cost \(\$ C\) per period to operate in any subsequent period (starting from period one). It cannot be operated in period zero, because it has not yet been built. In any period following a period in which the plant was operating, you will have output to sell. The sale of this output yields revenue of \(\$ R\). (You may assume that the retailing process incurs no additional costs. If you prefer, you can think of the revenue in any given period as being net revenue after the subtraction of any retailing costs incurred in that period.) In other words, if the plant is operated in period \(t\), then you will receive \(\$ R\) of revenue in period \((t+1)\). The plant will be worn out after exactly \(n\) periods of operation. Suppose that you decide to build the plant and operate it in all subsequent periods. Note that this means that the plant will begin operation in period one and it will be worn out after (at the end of) period \(n\). Thus the first period in which you will receive revenue is period two and the last period in which you will receive revenue is period \((n+1)\). Suppose that the per-period interest rate is denoted by \(i\) You may assume that this interest rate is fixed over the entire horizon of this project and that \(0<i<1\).

(a) What is the sequence of fixed costs for this project? What is the sequence of operational costs for this project? What is the sequence of revenues from this project? What is the stream of per-period profits for this project?

(b) What is the present value of the fixed cost sequence for this project?

(c) What is the present value of the operational cost sequence for this project?

(d) What is the present value of the revenue sequence for this project?

(e) What is the present value of the profit sequence for this project?

Solution

The timing of the receipts and outlays for this project is illustrated in Table One.

(a) The sequence of fixed costs is

\[ \left\{F_{0}, F_{1}, F_{2}, \cdots, F_{n+1}\right\}=\{F, 0,0, \cdots, 0\} \]

The sequence of operating costs is

\[ \left\{C_{0}, C_{1}, C_{2}, \cdots, C_{n}, C_{n+1}\right\}=\{0, C, C, \cdots, C, 0\} \]

The sequence of revenues for this project is

\[ \left\{R_{0}, R_{1}, R_{2}, R_{3}, \cdots, R_{n+1}\right\}=\{0,0, R, R, \cdots, R\} \]

The sequence of profits for this project is

\[ \left\{\pi_{0}, \pi_{1}, \pi_{2}, \pi_{3}, \cdots, \pi_{n}, \pi_{n+1}\right\}=\{-F,-C, R-C, R-C, \cdots, R-C, R\} \]

(b)

The present value of the fixed cost sequence for this project is

\[\begin{split} \begin{array}{l} P V(F) =\sum_{t=0}^{n+1} \frac{F_{t}}{(1+i)^{t}} \\ =\frac{F_{0}}{(1+i)^{0}}+\sum_{t=1}^{n+1} \frac{F_{t}}{(1+i)^{t}} \\ =\frac{F}{1}+\sum_{t=1}^{n+1} \frac{0}{(1+i)^{t}} \\ =F+0 \\ =F . \end{array} \end{split}\]

(c)

The present value of the operational cost sequence for this project is

\[\begin{split} \begin{array}{l} P V(C) =\sum_{t=0}^{n+1} \frac{C_{t}}{(1+i)^{t}} \\ =\frac{C_{0}}{(1+i)^{0}}+\sum_{t=1}^{n} \frac{C_{t}}{(1+i)^{t}}+\frac{C_{n+1}}{(1+i)^{n}} \\ =\frac{0}{1}+\frac{1}{(1+i)} \sum_{t=1}^{n} \frac{C}{(1+i)^{t-1}}+\frac{0}{(1+i)^{n}} \\ =0+\frac{1}{(1+i)} \sum_{k=0}^{n-1} \frac{C}{(1+i)^{k}}+0 \\ =\frac{C}{(1+i)} \sum_{k=0}^{n-1}\left(\frac{1}{1+i}\right)^{k} \\ =\frac{C}{(1+i)}\left\{\frac{1\left(1-\left(\frac{1}{1+i}\right)^{n}\right)}{1-\left(\frac{1}{1+i}\right)}\right\} \\ =\frac{C}{(1+i)}\left\{\frac{\left(1-\left(\frac{1}{1+i}\right)^{n}\right)}{\left(\frac{1+i-1}{1+i}\right)}\right\} \\ =\frac{C}{(1+i)}\left\{\frac{\left(1-\left(\frac{1}{1+i}\right)^{n}\right)}{\left(\frac{1+i - 1}{1+i}\right)}\right\} \\ =\frac{C}{(1+i)}\left\{\frac{\left(1-\left(\frac{1}{1+i}\right)^{n}\right)}{\left(\frac{i}{1+i}\right)}\right\} \\ =\frac{C(1+i)}{i(1+i)}\left(1-\left(\frac{1}{1+i}\right)^{n}\right) \\ =\left(\frac{C}{i}\right)\left(1-\left(\frac{1}{1+i}\right)^{n}\right) . \end{array} \end{split}\]

(d)

The present value of the revenue sequence for this project is

\[\begin{split} \begin{array}{l} P V(R)=\sum_{t=0}^{n+1} \frac{R_{t}}{(1+i)^{t}} \\ =\frac{R_{0}}{(1+i)^{0}}+\frac{R_{1}}{(1+i)^{1}}+\sum_{t=2}^{n+1} \frac{R_{t}}{(1+i)^{t}} \\ =\frac{0}{1}+\frac{0}{(1+i)}+\sum_{t=2}^{n+1} \frac{R}{(1+i)^{t}} \\ =0+0+\frac{1}{(1+i)^{2}} \sum_{t=2}^{n+1} \frac{R}{(1+i)^{t-2}} \\ =\frac{1}{(1+i)^{2}} \sum_{k=0}^{n-1} \frac{R}{(1+i)^{k}} \\ =\frac{R}{(1+i)^{2}} \sum_{k=0}^{n-1} \frac{1}{(1+i)^{k}} \\ =\frac{R}{(1+i)^{2}} \sum_{k=0}^{n-1}\left(\frac{1}{1+i}\right)^{k} \\ =\frac{R}{(1+i)^{2}}\left\{\frac{1\left(1-\left(\frac{1}{1+i}\right)^{n}\right)}{1-\left(\frac{1}{1+i}\right)}\right\} \\ =\frac{R}{(1+i)^{2}}\left\{\frac{\left(1-\left(\frac{1}{1+i}\right)^{n}\right)}{\left(\frac{1+i-1}{1+i}\right)}\right\} \\ =\frac{R}{(1+i)^{2}}\left\{\frac{\left(1-\left(\frac{1}{1+i}\right)^{n}\right)}{\left(\frac{1+i-1}{1+i}\right)}\right\} \\ =\frac{R}{(1+i)^{2}}\left\{\frac{\left(1-\left(\frac{1}{1+i}\right)^{n}\right)}{\left(\frac{i}{1+i}\right)}\right\} \\ =\frac{R(1+i)}{i(1+i)^{2}}\left(1-\left(\frac{1}{1+i}\right)^{n}\right) \\ =\left(\frac{R}{i(1+i)}\right)\left(1-\left(\frac{1}{1+i}\right)^{n}\right) . \end{array} \end{split}\]

(e)

The present value of the revenue sequence for this project is

\[\begin{split} \begin{array}{l} P V(\pi)=\sum_{t=0}^{n+1} \frac{\pi_{t}}{(1+i)^{t}} \\ =\frac{\pi_{0}}{(1+i)^{0}}+\frac{\pi_{1}}{(1+i)^{1}}+\sum_{t=2}^{n} \frac{\pi_{t}}{(1+i)^{t}}+\frac{\pi_{n}}{(1+i)^{n+1}} \\ =\frac{-F}{1}+\frac{-C}{(1+i)}+\sum_{t=2}^{n} \frac{(R-C)}{(1+i)^{t}}+\frac{R}{(1+i)^{n+1}} \\ =\left\{\frac{R}{(1+i)^{n+1}}-\frac{C}{(1+i)}-F\right\}+\frac{(R-C)}{(1+i)^{2}} \sum_{t=2}^{n} \frac{1}{(1+i)^{t-2}} \\ =\left\{\frac{R}{(1+i)^{n+1}}-\frac{C}{(1+i)}-F\right\}+\frac{(R-C)}{(1+i)^{2}} \sum_{t=2}^{n}\left(\frac{1}{1+i}\right)^{t-2} \\ =\left\{\frac{R}{(1+i)^{n+1}}-\frac{C}{(1+i)}-F\right\}+\frac{(R-C)}{(1+i)^{2}} \sum_{k=0}^{n-2}\left(\frac{1}{1+i}\right)^{k} \\ =\left\{\frac{R}{(1+i)^{n+1}}-\frac{C}{(1+i)}-F\right\}+\frac{(R-C)}{(1+i)^{2}}\left\{\frac{1\left(1-\left(\frac{1}{1+i}\right)^{n-1}\right)}{1-\left(\frac{1}{1+i}\right)}\right\} \\ =\left\{\frac{R}{(1+i)^{n+1}}-\frac{C}{(1+i)}-F\right\}+\frac{(R-C)}{(1+i)^{2}}\left\{\frac{\left(1-\left(\frac{1}{1+i}\right)^{n-1}\right)}{\left(\frac{1+i-1}{1+i}\right)}\right\} \\ =\left\{\frac{R}{(1+i)^{n+1}}-\frac{C}{(1+i)}-F\right\}+\frac{(R-C)}{(1+i)^{2}}\left\{\frac{\left(1-\left(\frac{1}{1+i}\right)^{n-1}\right)}{\left(\frac{i}{1+i}\right)}\right\} \\ =\left\{\frac{R}{(1+i)^{n+1}}-\frac{C}{(1+i)}-F\right\}+\frac{(R-C)(1+i)}{i(1+i)^{2}}\left\{1-\left(\frac{1}{1+i}\right)^{n-1}\right\} \\ =\left\{\frac{R}{(1+i)^{n+1}}-\frac{C}{(1+i)}-F\right\}+\frac{(R-C)}{i(1+i)}\left\{1-\left(\frac{1}{1+i}\right)^{n-1}\right\} . \end{array} \end{split}\]

Note that this can be rewritten as

\[\begin{split} \begin{array}{l} P V(\pi)=\frac{R}{(1+i)^{n+1}}-\frac{C}{(1+i)}-F+\frac{R}{i(1+i)}\left\{1-\left(\frac{1}{1+i}\right)^{n-1}\right\} -\frac{C}{i(1+i)}\left\{1-\left(\frac{1}{1+i}\right)^{n-1}\right\} \\ =\frac{R}{i(1+i)}\left\{\frac{i}{(1+i)^{n}}+1-\left(\frac{1}{1+i}\right)^{n-1}\right\} -\frac{C}{i(1+i)}\left\{i+1-\left(\frac{1}{1+i}\right)^{n-1}\right\}-F \\ =\frac{R}{i(1+i)}\left\{\frac{i}{(1+i)^{n}}+1-\frac{1}{(1+i)^{n-1}}\right\} -\frac{C}{i(1+i)}\left\{(1+i)-\frac{1}{(1+i)^{n-1}}\right\}-F \\ =\frac{R}{i(1+i)}\left\{\frac{i}{(1+i)^{n}}+\frac{(1+i)^{n}}{(1+i)^{n}}-\frac{(1+i)}{(1+i)^{n}}\right\} -\frac{C}{i}\left\{\frac{(1+i)}{(1+i)}-\frac{1}{(1+i)^{n}}\right\}-F \\ =\frac{R}{i(1+i)}\left\{\frac{i+(1+i)^{n}-(1+i)}{(1+i)^{n}}\right\} -\frac{C}{i}\left\{1-\frac{1}{(1+i)^{n}}\right\}-F \\ =\frac{R}{i(1+i)}\left\{\frac{i+(1+i)^{n}-1-i}{(1+i)^{n}}\right\} -\frac{C}{i}\left\{1-\frac{1}{(1+i)^{n}}\right\}-F \\ =\frac{R}{i(1+i)}\left\{\frac{(1+i)^{n}-1}{(1+i)^{n}}\right\}-\frac{C}{i}\left\{1-\frac{1}{(1+i)^{n}}\right\}-F \\ =\frac{R}{i(1+i)}\left\{\frac{(1+i)^{n}}{(1+i)^{n}}-\frac{1}{(1+i)^{n}}\right\}-\frac{C}{i}\left\{1-\frac{1}{(1+i)^{n}}\right\}-F \\ =\frac{R}{i(1+i)}\left\{1-\left(\frac{1}{1+i}\right)^{n}\right\}-\frac{C}{i}\left\{1-\left(\frac{1}{1+i}\right)^{n}\right\}-F \\ =P V(R)-P V(C)-P V(F) . \end{array} \end{split}\]