The first four terms in a sequence of numbers are given below. $T_1 = 5^2 - 18 = 7$ $T_2 = 6^2 - 22 = 14$ $T_3 = 7^2 - 26 = 23$ $T_4 = 8^2 - 30 = 34$

1. Extend the given pattern to find the fifth term.

$T_5=9^2-34=81-34=47$

1. Write the nth term using the pattern in the given sequence definitions.

From the given terms,

The squared number is $n+4$.

The numbers subtracted are $18,22,26,30$, increasing by $4$, so the nth number subtracted is
$18+4(n-1)=4n+14.$
Therefore,
$T_n=(n+4)^2-(4n+14).$

2. Expand and simplify the expression to show the required formula.

$T_n=(n+4)^2-(4n+14)$
$=n^2+8n+16-4n-14$
$=n^2+4n+2.$

1. Form an equation by equating the nth term to 962.

From part (b),
$T_n=n^2+4n+2$
For $962$ to be a term,
$n^2+4n+2=962$

2. Solve the quadratic equation for $n$ and decide whether it gives a valid term number.

$n^2+4n+2=962$
$n^2+4n-960=0$
$(n+32)(n-30)=0$
So $n=-32$ or $n=30$.

Since $n$ must be a positive integer, $n=30$ is valid.

1. Write expressions for $T_n$ and $T_{n-1}$.

From part (b),
$T_n=n^2+4n+2.$
Substitute $n-1$ for $n$:
$T_{n-1}=(n-1)^2+4(n-1)+2.$

2. Expand and simplify $T_{n-1}$.

$T_{n-1}=n^2-2n+1+4n-4+2$
$T_{n-1}=n^2+2n-1$

3. Subtract $T_{n-1}$ from $T_n$ and simplify.

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

1. Use the expression for the difference and explain why it is odd.

From part (d),
$T_n-T_{n-1}=2n+3.$
Since $2n$ is even for any integer $n$, $2n+3$ is even $+$ odd, which is odd. Hence the difference between consecutive terms is always odd.