The sequence $T_1, T_2, T_3, T_4,\dots$ has the following first four terms. $T_1 = 2^2 - 1 = 3$ $T_2 = 3^2 - 2 = 7$ $T_3 = 4^2 - 3 = 13$ $T_4 = 5^2 - 4 = 21$

1. Extend the given pattern to the 7th term and calculate it.

From the pattern,
$T_7=8^2-7=64-7=57.$

1. Generalise the given pattern to the $n$th term.

From the pattern,

$T_1=2^2-1=(1+1)^2-1$

$T_2=3^2-2=(2+1)^2-2$

$T_3=4^2-3=(3+1)^2-3$

So, replacing the term number by $n$,

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

1. Write expressions for two consecutive terms using the nth term.

From part (b),
$T_n=(n+1)^2-n$.

So
$T_{n+1}=((n+1)+1)^2-(n+1)=(n+2)^2-(n+1).$

2. Find and simplify the difference between consecutive terms.

$T_{n+1}-T_n=[(n+2)^2-(n+1)]-[(n+1)^2-n]$
$=n^2+4n+4-n-1-(n^2+2n+1-n)$
$=n^2+3n+3-(n^2+n+1)$
$=2n+2.$

3. Factorise the result and explain why it is always even.

$2n+2=2(n+1).$
Since $n$ is a positive integer, $n+1$ is an integer. Therefore $2(n+1)$ is a multiple of 2, so it is always even.