Question

430 number sequences (chapter 24)
opening problem
suppose we draw all the diagonals of an n - sided polygon. the first few cases are:
$n = 3$
0 diagonals
$n = 4$
2 diagonals
$n = 5$
5 diagonals
$n = 6$
9 diagonals
things to think about:
a consider the number of diagonals in each figure:
0 2 5 9
i can you describe the pattern to get from one number to the next?
ii can you hence predict the number of diagonals in a heptagon (7 - gon)?
draw a heptagon and all its diagonals to check your answer.
b consider the function $f(n)=\frac{n(n - 3)}{2}$.
what is the value of:
i $f(3)$
ii $f(4)$
iii $f(5)$
iv $f(6)$
v $f(7)$?
c can you write a formula for the number of diagonals in an n - sided polygon?
d can you explain why your formula gives the correct number of diagonals?
a
number sequences
a number sequence is an ordered list of numbers defined by a rule.
the numbers in a sequence are called the terms of the sequence.
the nth term or general term is written as $u_n$, so the sequence is $u_1,u_2,u_3,u_4,\ldots$
for example, consider this pattern of balls:
the first layer has just one blue ball, so $u_1 = 1$.
the second layer has three pink balls, so $u_2 = 3$.
the third layer has five black balls, so $u_3 = 5$.
the fourth layer has seven green balls, so $u_4 = 7$.
the pattern could be continued forever, generating a sequence of numbers.
there are many ways to describe this sequence:

• we can list the terms in order: 1, 3, 5, 7, 9, 11, 13, \ldots

the string of dots indicates that the pattern continues forever.

• we can write the sequence in words: “the sequence starts at 1 and increases by 2 each time.”

Explanation:

Step1: Analyze the pattern in part a - i

The differences between consecutive terms are: \(2 - 0=2\), \(5 - 2 = 3\), \(9 - 5=4\). The difference between the number of diagonals of an \((n + 1)\) - sided polygon and an \(n\) - sided polygon increases by 1 each time.

Step2: Predict number of diagonals for heptagon in part a - ii

Since the differences are increasing by 1, the next difference for \(n=7\) (heptagon) compared to \(n = 6\) is 5. So the number of diagonals of a heptagon is \(9+5 = 14\).

Step3: Calculate function values in part b

i. Calculate \(f(3)\)

Substitute \(n = 3\) into \(f(n)=\frac{n(n - 3)}{2}\), we get \(f(3)=\frac{3\times(3 - 3)}{2}=\frac{3\times0}{2}=0\).

ii. Calculate \(f(4)\)

Substitute \(n = 4\) into \(f(n)=\frac{n(n - 3)}{2}\), we get \(f(4)=\frac{4\times(4 - 3)}{2}=\frac{4\times1}{2}=2\).

iii. Calculate \(f(5)\)

Substitute \(n = 5\) into \(f(n)=\frac{n(n - 3)}{2}\), we get \(f(5)=\frac{5\times(5 - 3)}{2}=\frac{5\times2}{2}=5\).

iv. Calculate \(f(6)\)

Substitute \(n = 6\) into \(f(n)=\frac{n(n - 3)}{2}\), we get \(f(6)=\frac{6\times(6 - 3)}{2}=\frac{6\times3}{2}=9\).

v. Calculate \(f(7)\)

Substitute \(n = 7\) into \(f(n)=\frac{n(n - 3)}{2}\), we get \(f(7)=\frac{7\times(7 - 3)}{2}=\frac{7\times4}{2}=14\).

Step4: Answer part c

The formula for the number of diagonals in an \(n\) - sided polygon is \(f(n)=\frac{n(n - 3)}{2}\).

Step5: Explain the formula in part d

Each vertex of an \(n\) - sided polygon can be connected to \((n-3)\) other vertices to form diagonals (it cannot be connected to itself and the two adjacent vertices). There are \(n\) vertices, but we have double - counted each diagonal (since connecting vertex \(A\) to vertex \(B\) is the same as connecting vertex \(B\) to vertex \(A\)). So the number of diagonals is \(\frac{n(n - 3)}{2}\).

Answer:

a - i: The difference between the number of diagonals of an \((n + 1)\) - sided polygon and an \(n\) - sided polygon increases by 1 each time.
a - ii: 14
b - i: 0
b - ii: 2
b - iii: 5
b - iv: 9
b - v: 14
c: \(f(n)=\frac{n(n - 3)}{2}\)
d: Each vertex can be connected to \((n - 3)\) other vertices, and we divide by 2 to avoid double - counting.