Example 1: Given quadratic sequence is defined as $$ T_{k}=\sum_{n=0}^{k-1}(6 n-2) $$ 1. Show that the first 3 terms of the quadratic sequence are given by: $$ -2 ; 2 ; 12 ; \ldots \ldots $$

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To demonstrate that the first three terms of the quadratic sequence defined by $$ T_{k} = \sum_{n=0}^{k-1} (6n - 2) $$ are $$-2$$, $$2$$, and $$12$$, we'll calculate each term individually by evaluating the sum for $$ k = 1 $$, $$ k = 2 $$, and $$ k = 3 $$.

### Calculating $$ T_1 $$

For $$ k = 1 $$:
$$
T_1 = \sum_{n=0}^{0} (6n - 2)
$$
Since the upper limit is $$ n = 0 $$, the sum includes only one term ($$ n = 0 $$):
$$
T_1 = 6(0) - 2 = 0 - 2 = -2
$$

### Calculating $$ T_2 $$

For $$ k = 2 $$:
$$
T_2 = \sum_{n=0}^{1} (6n - 2)
$$
This sum includes two terms ($$ n = 0 $$ and $$ n = 1 $$):
$$
\begin{align*}
T_2 &= (6(0) - 2) + (6(1) - 2) \
&= -2 + (6 - 2) \
&= -2 + 4 \
&= 2
\end{align*}
$$

### Calculating $$ T_3 $$

For $$ k = 3 $$:
$$
T_3 = \sum_{n=0}^{2} (6n - 2)
$$
This sum includes three terms ($$ n = 0 $$, $$ n = 1 $$, and $$ n = 2 $$):
$$
\begin{align*}
T_3 &= (6(0) - 2) + (6(1) - 2) + (6(2) - 2) \
&= -2 + 4 + (12 - 2) \
&= -2 + 4 + 10 \
&= 12
\end{align*}
$$

### Summary of the First Three Terms
$$
T_1 = -2 \
T_2 = 2 \
T_3 = 12 \
$$
Thus, the first three terms of the quadratic sequence are $$-2$$, $$2$$, and $$12$$.
The first three terms of the sequence are -2, 2, and 12.

Example 1: Given quadratic sequence is defined as \( T_{k}=\sum_{n=0}^{k-1}(6 n-2) \) 1. Show that the first 3 terms of the quadratic sequence are given by: \( -2 ; 2 ; 12 ; \ldots \ldots \)

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