Solutions#

Question 1#

1. Using recursion, obtain the first 10 terms of the following sequences:

1. \(\left\{\begin{array}{l}a_1 = 1,\\a_n = 3a_{n - 1}, n > 1\end{array}\right.\)

def get_sequence_a(n):
    """
    Return the sequence a.
    """
    if n == 1:
        return 1
    return 3 * get_sequence_a(n - 1)


[get_sequence_a(n) for n in range(1, 11)]

[1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683]

2. \(\left\{\begin{array}{l}b_1 = 3,\\b_n = 6b_{n - 1}, n > 1\end{array}\right.\)

def get_sequence_b(n):
    """
    Return the sequence b.
    """
    if n == 1:
        return 3
    return 6 * get_sequence_b(n - 1)


[get_sequence_b(n) for n in range(1, 11)]

[3, 18, 108, 648, 3888, 23328, 139968, 839808, 5038848, 30233088]

3. \(\left\{\begin{array}{l}c_1 = 3,\\c_n = 6c_{n - 1} + 3, n > 1\end{array}\right.\)

def get_sequence_c(n):
    """
    Return the sequence c.
    """
    if n == 1:
        return 3
    return 6 * get_sequence_c(n - 1) + 3


[get_sequence_c(n) for n in range(1, 11)]

[3, 21, 129, 777, 4665, 27993, 167961, 1007769, 6046617, 36279705]

4. \(\left\{\begin{array}{l}d_0 = 3,\\d_n = \sqrt{d_{n - 1}} + 3, n > 0\end{array}\right.\)

import sympy as sym


def get_sequence_d(n):
    """
    Return the sequence c.
    """
    if n == 0:
        return 3
    return sym.sqrt(get_sequence_d(n - 1)) + 3


[get_sequence_d(n) for n in range(1, 11)]

[sqrt(3) + 3,
 sqrt(sqrt(3) + 3) + 3,
 sqrt(sqrt(sqrt(3) + 3) + 3) + 3,
 sqrt(sqrt(sqrt(sqrt(3) + 3) + 3) + 3) + 3,
 sqrt(sqrt(sqrt(sqrt(sqrt(3) + 3) + 3) + 3) + 3) + 3,
 sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(3) + 3) + 3) + 3) + 3) + 3) + 3,
 sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(3) + 3) + 3) + 3) + 3) + 3) + 3) + 3,
 sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(3) + 3) + 3) + 3) + 3) + 3) + 3) + 3) + 3,
 sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(3) + 3) + 3) + 3) + 3) + 3) + 3) + 3) + 3) + 3,
 sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(3) + 3) + 3) + 3) + 3) + 3) + 3) + 3) + 3) + 3) + 3]

We could use a sqrt from a different library. Choosing to use sympy as it ensures the result is exact although not necessarily readable. Here is an approximate approach:

import math


def get_sequence_d(n):
    """
    Return the sequence c.
    """
    if n == 0:
        return 3
    return math.sqrt(get_sequence_d(n - 1)) + 3


[get_sequence_d(n) for n in range(10)]

[3,
 4.732050807568877,
 5.175327747161075,
 5.2749346687676715,
 5.296722592906612,
 5.301460969233807,
 5.302490167022176,
 5.302713652850084,
 5.302762178960321,
 5.3027727154368325]

Question 2#

  1. Using recursion, obtain the first 5 terms of the Fibonacci sequence:

\[\begin{split} \left\{ \begin{array}{l} a_0 = 0,\\ a_1 = 1,\\ a_n = a_{n - 1} + a_{n - 2}, n \geq 2\end{array}\right. \end{split}\]
def get_fibonacci(n):
    """
    Return the nth term of the Fibonacci sequence
    """
    if n == 0:
        return 0
    if n == 1:
        return 1
    return get_fibonacci(n - 1) + get_fibonacci(n - 2)


[get_fibonacci(n) for n in range(10)]

[0, 1, 1, 2, 3, 5, 8, 13, 21, 34]

Question 3#

3. A 40 year building programme for new houses began in Oldtown in the year 1951 (Year 1) and finished in 1990 (Year 40).

The number of houses built each year form an arithmetic sequence with first term \(a\) and common difference \(d\).

Given that 2400 new houses were built in 1960 and 600 new houses were built in 1990, find:

1. The value of \(d\).

An arithmetic sequence with first term \(a\) and common difference \(d\) is a sequence of the form:

\[\begin{split} \left\{ \begin{array}{l} a_1 = a,\\ a_n = a_{n - 1} + d \end{array} \right. \end{split}\]

We will write a function to express this:

def get_arithmetic_sequence(n, first_term, common_difference):
    """
    Return the nth term of an arithmetic sequence with give first_term and
    common common_difference.
    """
    if n == 1:
        return first_term
    return (
        get_arithmetic_sequence(n - 1, first_term, common_difference)
        + common_difference
    )

We know that \(a_{10}=2400\) and \(a_{40}=600\) we can write down equations that represent this:

a = sym.Symbol("a")
d = sym.Symbol("d")

a_10_equation = sym.Eq(
    get_arithmetic_sequence(n=10, first_term=a, common_difference=d), 2400
)
a_10_equation
\[\displaystyle a + 9 d = 2400\]
a_40_equation = sym.Eq(
    get_arithmetic_sequence(n=40, first_term=a, common_difference=d), 600
)
a_40_equation
\[\displaystyle a + 39 d = 600\]

We will solve the first equation for \(a\):

sym.solveset(a_10_equation, a)
\[\displaystyle \left\{2400 - 9 d\right\}\]

We substitute this in to the other equation and solve it for \(d\):

sym.solveset(a_40_equation.subs({a: 2400 - 9 * d}), d)
\[\displaystyle \left\{-60\right\}\]

2. The value of \(a\).

We can substitute that value for \(d\) back in to the expression for \(a\):

(2400 - 9 * d).subs({d: -60})
\[\displaystyle 2940\]

3. The total number of houses built in Oldtown over 40 years.

sum(
    get_arithmetic_sequence(n=n, first_term=2940, common_difference=-60)
    for n in range(1, 41)
)

70800

Question 4#

4. A sequence is given by:

\[\begin{split} \left\{\begin{array}{l} x_1 = 1\\ x_{n + 1}= x_n(p + x_n), n > 1 \end{array}\right. \end{split}\]

for \(p\ne0\).

1. Find \(x_2\) in terms of \(p\).

We start by defining a function:

def get_sequence(n, p):
    """
    Return the nth term of the sequence x_n for a given value of p
    """
    if n == 1:
        return 1
    return get_sequence(n - 1, p) * (p + get_sequence(n - 1, p))

Using this we can answer the question:

p = sym.Symbol("p")

x_2 = get_sequence(n=2, p=p)
x_2
\[\displaystyle p + 1\]

2. Show that \(x_3=1+3p+2p^2\).

p = sym.Symbol("p")

x_3 = get_sequence(n=3, p=p)
sym.expand(x_3)
\[\displaystyle 2 p^{2} + 3 p + 1\]

3. Given that \(x_3=1\), find the value of \(p\)

equation = sym.Eq(x_3, 1)
sym.solveset(equation, p)
\[\displaystyle \left\{- \frac{3}{2}, 0\right\}\]

As \(p\ne0\) this gives us that \(p=-\frac{3}{2}\).