Aug 3, 2016

The ${n}^{\text{th}}$ term is given by $\left(n - 1\right) \left(n + 1\right)$

#### Explanation:

Examining the sequence, we find that the difference between successive terms increase by $2$, starting with an increase of $3$. That is, we have

${a}_{0} = 0$
${a}_{1} = 0 + 3$
${a}_{2} = 0 + 3 + 5$
${a}_{3} = 0 + 3 + 5 + 7$
$\ldots$

As the difference increases by $2$ each time, we can see that for each $k \ge 1$, the difference between ${a}_{k - 1}$ and ${a}_{k}$ is $3 + 2 \left(k - 1\right)$. Thus, noting that the ${n}^{\text{th}}$ term is ${a}_{n - 1}$ because we started from ${a}_{0}$, we have

${a}_{n - 1} = 0 + 3 + 5 + \ldots + \left(3 + 2 \left(\left(n - 1\right) - 1\right)\right)$

$= \left(3 + 2 \left(0\right)\right) + \left(3 + 2 \left(1\right)\right) + \ldots + \left(3 + 2 \left(n - 2\right)\right)$

$= \left(3 + 3 + \ldots + 3\right) + \left(2 \left(0\right) + 2 \left(1\right) + \ldots + 2 \left(n - 2\right)\right)$

$= 3 \left(n - 1\right) + 2 \left(1 + 2 + \ldots + \left(n - 2\right)\right)$

$= 3 \left(n - 1\right) + 2 \left(\frac{\left(n - 2\right) \left(n - 1\right)}{2}\right)$

(The above comes from that the sum of all successive integers from $1$ to $k$ is given by $\frac{k \left(k + 1\right)}{2}$, with $k = n - 2$ in this case)

$= 3 \left(n - 1\right) + \left(n - 2\right) \left(n - 1\right)$

$= \left(n - 1\right) \left(n + 1\right)$

Checking we find that this matches up with the given values. E.g. the ${3}^{\text{rd}}$ term is $8 = 2 \cdot 4 = \left(3 - 1\right) \left(3 + 1\right)$