Quadratic Sequences

Write down the next two terms and determine an equation for the \(n^{\text{th}}\) term of the sequence \(\text{5}\); \(\text{12}\); \(\text{23}\); \(\text{38}\); \(\ldots\)

Find the first differences between the terms

Find the second differences between the terms

So there is a common second difference of \(\text{4}\). We can therefore conclude that this is a quadratic sequence of the form \(T_n = an^2 + bn + c\).

Continuing the sequence, the next first differences will be:

Finding the next two terms in the sequence

The next two terms will be:

So the sequence will be: \(\text{5}\); \(\text{12}\); \(\text{23}\); \(\text{38}\); \(\text{57}\); \(\text{80}\); \(\ldots\)

Determine the general term for the sequence

To find the values of \(a\), \(b\) and \(c\) for \(T_n = an^2 + bn + c\) we look at the first \(\text{3}\) terms in the sequence:

\begin{align*} n=1: T_1 &= a + b + c \\ n=2: T_2 &= 4a + 2b + c \\ n=3: T_3 &= 9a + 3b + c \end{align*}

We solve a set of simultaneous equations to determine the values of \(a\), \(b\) and \(c\)

We know that \(T_1 = 5\), \(T_2 = 12\) and \(T_3 = 23\)

\begin{align*} a + b + c &= 5 \\ 4a + 2b + c &= 12 \\ 9a + 3b + c &= 23 \end{align*}\begin{align*} T_2 - T_1 &= 4a + 2b + c - (a + b + c) \\ 12 - 5 &= 4a + 2b + c - a - b - c \\ 7 &= 3a + b \qquad \ldots (1) \end{align*}\begin{align*} T_3 - T_2 &= 9a + 3b + c - (4a + 2b + c) \\ 23 - 12 &= 9a + 3b + c - 4a - 2b - c \\ 11 &= 5a + b \qquad \ldots (2) \end{align*}\begin{align*} (2)-(1) &= 5a + b - (3a + b) \\ 11 - 7 &= 5a + b - 3a - b \\ 4 &= 2a \\ \therefore a &= 2 \end{align*}\begin{align*} \text{Using equation } (1): \quad 3(2) + b &= 7 \\ \therefore b &= 1 \\ \text{And using } \quad a + b + c &= 5 \\ 2 + 1 + c &= 5 \\ \therefore c &= 1 \end{align*}

Write the general term for the sequence

\[T_n = 2n^2 + n + 2\]

Consider the following sequence:

\[3; 6; 10; 15; 21; \ldots\]

1. Determine the general term (\(T_n\)) for the sequence.
2. Is this a linear or a quadratic sequence?
3. Plot a graph of \(T_n\) vs \(n\).

Determine the first and second differences

We see that the first differences are not constant and form the sequence \({3; 4; 5; \ldots}\) and that there is a common second difference of \(\text{1}\). Therefore the sequence is quadratic and has a general term of the form \(T_n = an^2 + bn + c\).

Determine the general term \(T_n\)

To find the values of \(a\), \(b\) and \(c\) for \(T_n = an^2 + bn + c\) we look at the first \(\text{3}\) terms in the sequence:

\begin{align*} n=1: T_1 &= a + b + c \\ n=2: T_2 &= 4a + 2b + c \\ n=3: T_3 &= 9a + 3b + c \end{align*}

We solve this set of simultaneous equations to determine the values of \(a\), \(b\) and \(c\). We know that \(T_1 = 3\), \(T_2 = 6\) and \(T_3 = 10\).

\begin{align*} a + b + c &= 3 \\ 4a + 2b + c &= 6 \\ 9a + 3b + c &= 10 \end{align*}\begin{align*} T_2 - T_1 &= 4a + 2b + c - (a + b + c) \\ 6 - 3 &= 4a + 2b + c - a - b - c \\ 3 &= 3a + b \qquad \ldots (1) \end{align*}\begin{align*} T_3 - T_2 &= 9a + 3b + c - (4a + 2b + c) \\ 10 - 6 &= 9a + 3b + c - 4a - 2b - c \\ 4 &= 5a + b \qquad \ldots (2) \end{align*}\begin{align*} (2)-(1) &= 5a + b - (3a + b) \\ 4 - 3 &= 5a + b - 3a - b \\ 1 &= 2a \\ \therefore a &= \cfrac{1}{2} \end{align*}\begin{align*} \text{Using equation }(1): \quad 3\left(\cfrac{1}{2}\right) + b &= 3 \\ \therefore b &= \cfrac{3}{2} \\ \text{And using } \quad a + b + c &= 3 \\ \cfrac{1}{2} + \cfrac{3}{2} + c &= 3 \\ \therefore c &= 1 \end{align*}

Therefore the general term for the sequence is \(T_n = \cfrac{1}{2}n^2 + \cfrac{3}{2}n + 1\).

Plot a graph of \(T_n\) vs \(n\)

Use the general term for the sequence, \(T_n = \cfrac{1}{2}n^2 + \cfrac{3}{2}n + 1\), to complete the table.

\(n\)	\(\text{1}\)	\(\text{2}\)	\(\text{3}\)	\(\text{4}\)	\(\text{5}\)	\(\text{6}\)	\(\text{7}\)	\(\text{8}\)	\(\text{9}\)	\(\text{10}\)
\(T_n\)	\(\text{3}\)	\(\text{6}\)	\(\text{10}\)	\(\text{15}\)	\(\text{21}\)	\(\text{28}\)	\(\text{36}\)	\(\text{45}\)	\(\text{55}\)	\(\text{66}\)

Use the table to plot the graph:

In this case it would not be accurate to join these points, since \(n\) indicates the position of a term in a sequence and can therefore only be a positive integer. We can, however, see that the plot of the points lies in the shape of a parabola.

In the first stage of the soccer event at the Olympic Games, there are teams from four different countries in each group. Each country in a group must play every other country in the group once.

1. How many matches will be played in each group in the first stage of the event?
2. How many matches would be played if there are \(\text{5}\) teams in each group?
3. How many matches would be played if there are \(\text{6}\) teams in each group?
4. Determine the general formula of the sequence.

Determine the number of matches played if there are \(\text{4}\) teams in a group

Let the teams from four different countries be \(A\), \(B\), \(C\) and \(D\).

teams in a group	matches played
\(A\)	\(AB, AC, AD\)
\(B\)	\(BC, BD\)
\(C\)	\(CD\)
\(D\)
\(\text{4}\)	\(3+2+1=6\)

\(AB\) means that team \(A\) plays team \(B\) and \(BA\) would be the same match as \(AB\). So if there are four different teams in a group, each group plays \(\text{6}\) matches.

Determine the number of matches played if there are \(\text{5}\) teams in a group

Let the teams from five different countries be \(A, B, C, D\) and \(E\).

teams in a group	matches played
\(A\)	\(AB, AC, AD, AE\)
\(B\)	\(BC, BD, BE\)
\(C\)	\(CD, CE\)
\(D\)	\(DE\)
\(E\)
\(\text{5}\)	\(4+3+2+1=10\)

So if there are five different teams in a group, each group plays \(\text{10}\) matches.

Determine the number of matches played if there are \(\text{6}\) teams in a group

Let the teams from six different countries be \(A, B, C, D, E\) and \(F\).

teams in a group	matches to be played
\(A\)	\(AB, AC, AD, AE, AF\)
\(B\)	\(BC, BD, BE, BF\)
\(C\)	\(CD, CE, CF\)
\(D\)	\(DE, DF\)
\(E\)	\(EF\)
\(F\)
\(\text{5}\)	\(5+4+3+2+1=15\)

So if there are six different teams in a group, each group plays \(\text{15}\) matches.

We continue to increase the number of teams in a group and find that a group of \(\text{7}\) teams plays \(\text{21}\) matches and a group of \(\text{8}\) teams plays \(\text{28}\) matches.

Consider the sequence

We examine the sequence to determine if it is linear or quadratic:

We see that the first differences are not constant and that there is a common second difference of \(\text{1}\). Therefore the sequence is quadratic and has a general term of the form \(T_n = an^2 + bn + c\).

Determine the general term \(T_n\)

To find the values of \(a\), \(b\) and \(c\) for \(T_n = an^2 + bn + c\) we look at the first \(\text{3}\) terms in the sequence:

\begin{align*} n=1: T_1 &= a + b + c \\ n=2: T_2 &= 4a + 2b + c \\ n=3: T_3 &= 9a + 3b + c \end{align*}

We solve a set of simultaneous equations to determine the values of \(a\), \(b\) and \(c\). We know that \(T_1 = 6\), \(T_2 = 10\) and \(T_3 = 15\)

\begin{align*} a + b + c &= 6 \\ 4a + 2b + c &= 10 \\ 9a + 3b + c &= 15 \end{align*}\begin{align*} T_2 - T_1 &= 4a + 2b + c - (a + b + c) \\ 10 - 6 &= 4a + 2b + c - a - b - c \\ 4 &= 3a + b \qquad \ldots (1) \end{align*}\begin{align*} T_3 - T_2 &= 9a + 3b + c - (4a + 2b + c) \\ 15- 10 &= 9a + 3b + c - 4a - 2b - c \\ 5 &= 5a + b \qquad \ldots (2) \end{align*}\begin{align*} (2)-(1) &= 5a + b - (3a + b) \\ 5-4 &= 5a + b - 3a - b \\ 1 &= 2a \\ \therefore a &= \cfrac{1}{2} \end{align*}\begin{align*} \text{Using equation } (1): \quad 3\left(\cfrac{1}{2}\right) + b &= 4 \\ \therefore b &= \cfrac{5}{2} \\ \text{And using } a + b + c &= 6 \\ \cfrac{1}{2} + \cfrac{5}{2} + c &= 6 \\ \therefore c &= 3 \end{align*}

Therefore the general term for the sequence is \(T_n = \cfrac{1}{2}n^2 + \cfrac{5}{2}n + 3\).