The figure shows the first three figures of a sequence formed by sticks of the same size.

1. Substitute the figure number into the given pattern and evaluate.

For Figure 5,
$(5+1)^2-1=6^2-1=36-1=35$

1. Generalise the given pattern by replacing the figure number with $n$.

From the table,

Figure $1$: $(1+1)^2-1$

Figure $2$: $(2+1)^2-1$

Figure $3$: $(3+1)^2-1$

So for Figure $n$, the number of squares is

$(n+1)^2-1.$

1. Write expressions for the number of squares in two consecutive figures.

From part (ii), the number of squares in Figure $n$ is
$(n+1)^2-1.$
So the number of squares in Figure $n+1$ is
$(n+2)^2-1.$

2. Add the two expressions and show the sum is odd.

Sum of the number of squares:
$[(n+1)^2-1]+[(n+2)^2-1]$
$=n^2+2n+n^2+4n+3$
$=2n^2+6n+3$
$=2(n^2+3n+1)+1.$
This is of the form $2k+1$, so it is always odd.

1. Decide whether $4n+3$ can be a multiple of 7.

For Sequence 1, the $n^{th}$ term is $4n+3$.

Test values of $n$:

When $n=1$,
$4(1)+3=7,$
which is a multiple of 7.

When $n=2$,
$4(2)+3=11,$
which is not a multiple of 7.

So the terms are multiples of 7 only for some values of $n$.

1. Determine whether $7n+1$ is ever a multiple of 7.

$7n$ is always a multiple of $7$.

So $7n+1$ is always $1$ more than a multiple of $7$, and hence is never a multiple of $7$.

1. Identify that the nth term is a multiple of 7 for every positive integer value of $n$.

Sequence 3 has nth term
$14n=7(2n).$
Since $14n$ is divisible by $7$ for all values of $n$, the sequence is always made up of multiples of $7$.

1. Rewrite the nth term to identify when it is a multiple of 7.

For Sequence 4,
$8n-1=7n+(n-1).$
This is a multiple of 7 when $n-1$ is a multiple of 7, for example when $n=1,8,15,\ldots$

2. Decide the correct classification.

For example:
$n=1:\quad 8(1)-1=7,$
which is a multiple of 7.

But
$n=2:\quad 8(2)-1=15,$
which is not a multiple of 7.

So the terms are sometimes multiples of 7.