Study the pattern below.

1. Recognise the pattern for the bottom two cells.

From Figures 1 to 4,

bottom-left values are $1,2,3,4$, so in Figure $11$,
$b=11.$

bottom-right values are $1,4,9,16$, so these are squares:
$c=11^2=121.$

2. Recognise the pattern for the top cell and find its value in Figure 11.

Top values are $2,6,12,20$.

Each is the figure number multiplied by one more than itself:
$1\times2=2,\quad 2\times3=6,\quad 3\times4=12,\quad 4\times5=20.$

So in Figure $11$,
$a=11\times12=132.$

1. Identify the rule for the top cell in Figure $n$.

The top values are
$2,\ 6,\ 12,\ 20,\ldots$
These can be written as
$1\times2,\ 2\times3,\ 3\times4,\ 4\times5,\ldots$
So for Figure $n$,
$a=n(n+1).$

1. Use the expression for $a$ from the pattern.

From part (b),

$a=n(n+1).$

This means $a$ is the product of two consecutive integers, $n$ and $n+1$.

2. Explain why the product is always even.

For any integer value of $n$, one of the two consecutive integers $n$ and $n+1$ must be even.

Therefore, $n(n+1)$ is divisible by $2$, so $a$ is even.