QUESTION 2: Arithmetic and Geometric Sequences and Series. QUESTION 3: Quadratic Number Patterns and Series.

$$300 - 18 = (n-1)6$$ $$282 = 6(n-1)$$ $$\frac{282}{6} = n-1$$ $$47 = n-1$$ $$n = 47 + 1$$ $$n = 48$$

The number of terms in the series is $\boxed{48}$.

2.1.2 Calculate the sum of this series.

$$S_{48} = 24(318)$$ $$S_{48} = 7632$$

The sum of this series is $\boxed{7632}$.

2.1.3 Calculate the sum of all the whole numbers up to and including 300 that are NOT divisible by 6.

The sum of the first $N$ whole numbers is $S_N = \frac{N(N+1)}{2}$. Here $N=300$.

$$S_{300} = \frac{300(300+1)}{2}$$ $$S_{300} = \frac{300 \times 301}{2}$$ $$S_{300} = 150 \times 301$$ $$S_{300} = 45150$$

These numbers form an arithmetic series: $6, 12, 18, \dots, 300$.
First term $a' = 6$. Common difference $d' = 6$. Last term $T_k' = 300$.
Find the number of terms $k$:

$$T_k' = a' + (k-1)d'$$ $$300 = 6 + (k-1)6$$ $$294 = 6(k-1)$$ $$\frac{294}{6} = k-1$$ $$49 = k-1$$ $$k = 50$$

Now, calculate the sum $S_k'$:

$$S_k' = \frac{k}{2}(a' + T_k')$$ $$S_{50}' = \frac{50}{2}(6 + 300)$$ $$S_{50}' = 25(306)$$ $$S_{50}' = 7650$$

Required Sum = (Sum of all numbers up to 300) - (Sum of numbers divisible by 6 up to 300)

$$\text{Required Sum} = 45150 - 7650$$ $$\text{Required Sum} = 37500$$

The sum of all whole numbers up to and including 300 that are NOT divisible by 6 is $\boxed{37500}$.

Question 2.2: Geometric Sequence

The first three terms of an infinite geometric sequence are 16, 8, and 4.
The first term is $a = 16$.
The common ratio is $r = \frac{8}{16} = \frac{1}{2}$.

2.2.1 Determine the $n^{th}$ term of the sequence.

$$T_n = 16 \left(\frac{1}{2}\right)^{n-1}$$

The $n^{th}$ term of the sequence is $\boxed{T_n = 16 \left(\frac{1}{2}\right)^{n-1}}$.

2.2.2 Determine all possible values of $n$ for which the sum of the first $n$ terms of this sequence is greater than 31.

We are given $S_n > 31$.

$$\frac{16\left(1 - \left(\frac{1}{2}\right)^n\right)}{1 - \frac{1}{2}} > 31$$

$$\frac{16\left(1 - \left(\frac{1}{2}\right)^n\right)}{\frac{1}{2}} > 31$$ $$32\left(1 - \left(\frac{1}{2}\right)^n\right) > 31$$ $$1 - \left(\frac{1}{2}\right)^n > \frac{31}{32}$$ $$1 - \frac{31}{32} > \left(\frac{1}{2}\right)^n$$ $$\frac{1}{32} > \left(\frac{1}{2}\right)^n$$

$$\left(\frac{1}{2}\right)^5 > \left(\frac{1}{2}\right)^n$$

Since the base $\frac{1}{2}$ is between 0 and 1, the inequality sign flips when comparing the exponents.

$$5 < n$$

Since $n$ represents the number of terms, it must be an integer.
The possible values of $n$ are $\boxed{n > 5, \text{ where } n \in \mathbb{Z}}$.

2.2.3 Calculate the sum to infinity of this sequence.

$$S_\infty = \frac{16}{\frac{1}{2}}$$ $$S_\infty = 16 \times 2$$ $$S_\infty = 32$$

The sum to infinity of this sequence is $\boxed{32}$.