下面给出了一个数列的前四项。 $T_1 = 3^2 + 5 = 14$ $T_2 = 4^2 + 8 = 24$ $T_3 = 5^2 + 11 = 36$ $T_4 = 6^2 + 14 = 50$

1. Extend the given pattern to the fifth term and evaluate it.

From the given terms,
$T_1=3^2+5,\quad T_2=4^2+8,\quad T_3=5^2+11,\quad T_4=6^2+14.$
So
$T_5=7^2+17=49+17=66.$

1. Explain why the two parts of each term always add to an even number.

In each term, the base being squared and the number added have the same parity:

$3^2+5$ is odd $+$ odd, $4^2+8$ is even $+$ even, and this pattern continues.

An odd square is odd and an even square is even. Hence each $T_n$ is either odd $+$ odd or even $+$ even, both of which are even.

1. Express the two parts of the pattern in terms of $n$.

From the given terms,

The squared number is $2$ more than the term number, so it is $n+2$.

The number added increases by $3$ each time: $5,8,11,14,\ldots$, so the nth number added is
$5+3(n-1)=3n+2.$

2. Form $T_n$ and simplify to show the required expression.

Hence
$T_n=(n+2)^2+(3n+2).$
Expanding and simplifying,
$T_n=n^2+4n+4+3n+2=n^2+7n+6.$

1. Write expressions for $T_{p+1}$ and $T_p$ using the nth term formula.

From part (c),
$T_n=n^2+7n+6.$
So
$T_{p+1}=(p+1)^2+7(p+1)+6$
and
$T_p=p^2+7p+6.$

2. Find $T_{p+1}-T_p$ and simplify.

$T_{p+1}-T_p=[(p+1)^2+7(p+1)+6]-(p^2+7p+6)$
$=p^2+2p+1+7p+7+6-p^2-7p-6$
$=2p+8.$

1. Use the expression for the difference between consecutive terms.

From part (d),

$T_{p+1}-T_p=2p+8.$

2. Explain why this difference cannot be 4.

Since $p$ is a positive integer, $p\geq 1$.

Hence

$2p+8\geq 2(1)+8=10.$

So $T_{p+1}-T_p$ cannot be $4$.