一个数列的前四项如下所示。 $T_1 = (1\times 3) + 8 = 11$ $T_2 = (2\times 4) + 12 = 20$ $T_3 = (3\times 5) + 16 = 31$ $T_4 = (4\times 6) + 20 = 44$

1. Use the observed pattern to form and calculate the fifth term

From the pattern,

$T_5 = (5\times 7)+24$

$=35+24$

$=59$

1. Express the observed pattern for the $n^{\text{th}}$ term

From the given terms,
$T_1=(1\times3)+8=(1)(1+2)+4(1+1)$
$T_2=(2\times4)+12=(2)(2+2)+4(2+1)$
$T_3=(3\times5)+16=(3)(3+2)+4(3+1)$
So, in general,
$T_n=n(n+2)+4(n+1).$

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

$T_n=n(n+2)+4(n+1)$
$=n^2+2n+4n+4$
$=n^2+6n+4$

1. Find an expression for $T_{p-1}$ using the nth term formula

From part (b),
$T_n=n^2+6n+4$
Substitute $n=p-1$:
$T_{p-1}=(p-1)^2+6(p-1)+4$
$=p^2-2p+1+6p-6+4$
$=p^2+4p-1$

2. Find an expression for $T_p$ using the nth term formula

Substitute $n=p$:
$T_p=p^2+6p+4$

3. Add the two consecutive terms and simplify

$T_{p-1}+T_p=(p^2+4p-1)+(p^2+6p+4)$
$=2p^2+10p+3$

1. Form a quadratic equation using the sum from part (c)

From part (c),
$T_{p-1}+T_p=2p^2+10p-1.$
Given that the sum is $303$,
$2p^2+10p-1=303$
$2p^2+10p-304=0$
$p^2+5p-152=0$

2. Solve the quadratic equation and select the valid value of $p$

$p^2+5p-152=0$
$(p+16)(p-11)=0$
So,
$p=-16 \quad \text{or} \quad p=11.$
Since $p$ is a term number, $p$ must be positive.
$p=11$