Consider the sequence: $T_1 = 1 = 1$ $T_2 = 1 + 2 = 3$ $T_3 = 1 + 2 + 3 = 6$ $T_4 = 1 + 2 + 3 + 4 = 10$ $T_5 = 1 + 2 + 3 + 4 + 5 = 15$

1. Write the next term in the same form and evaluate it.

Since $T_5 = 1 + 2 + 3 + 4 + 5 = 15$,

$T_6 = 1 + 2 + 3 + 4 + 5 + 6 = 21$.

1. Use the given nth-term formula to find $T_{100}$.

Given $T_n=\dfrac{n(n+1)}{2}$,

$T_{100}=\dfrac{100(100+1)}{2}=\dfrac{100\times101}{2}=5050$

1. Express the given sum as a multiple of the triangular number $T_{100}$.

$3+6+9+12+\ldots+300 = 3(1+2+3+4+\ldots+100)=3T_{100}$

2. Use the value of $T_{100}$ or the given formula to calculate the sum.

Using $T_{100}=5050$,

$3T_{100}=3(5050)=15150$

1. Find the common difference and form the nth term.

The sequence is $1, 4, 7, 10, \ldots$

Common difference $= 3$.

$T_n = 1 + (n-1)(3) = 3n - 2$

1. Substitute $n=80$ into the nth term of the sequence.

From part (i), the nth term is $3n-2$.

For the 80th term,
$3(80)-2=240-2=238$

1. Find the smallest four-digit term in the sequence.

The nth term is $3n-2$.

For the smallest four-digit term,
$3n-2\geq 1000$
$3n\geq 1002$
$n\geq 334$

So the smallest four-digit term is
$3(334)-2=1000.$

2. Find the largest four-digit term in the sequence.

For the largest four-digit term,
$3n-2\leq 9999$
$3n\leq 10001$
$n\leq 3333\dfrac{2}{3}$

Since $n$ is a positive integer, the largest possible value is $n=3333$.

So the largest four-digit term is
$3(3333)-2=9997.$