Guess explicit formula for sequence and then prove it is correct just trying to work through the following problem but have come to point where I don't know what to do.
Info: Let $a_0, a_1, a_2, ...$ be the sequence defined recursively as $a_0 = 0, a_k = k + a_{k-1}$ for each integer $k\ge 1$.
Question is: Using this information write out the first 6 terms of the sequence, guess the explicit formula and then prove your guess is correct.
So I've done as follows:
First 6 terms: $a_0 = 0, a_1 = 1, a_2 = 3, a_3 = 6, a_4 = 10, a_5 = 15, a_6 = 21$
Explicit formula guess: $ a_n = \frac {n^2 + n}{2}$
Proof:
Using the guessed explicit formula $a_n = \frac {n^2 + n}{2}$ for each integer $n \ge 0$ we see that
$a_0 = \frac {0^2 + 0}{2} = 0$ so we have the correct initial condition.
Let $a_0, a_1, a_2, ...$ be the sequence defined as $a_0 = 0$ and $a_k = k + a_{k-1}$ for each integer $k \ge 1$
Then, let P(n) be the predicate $a_n = \frac {n^2 + n}{2}$ for each integer $n \ge 0$
Basis step: $a_0 = 0$ from the recursive definition and $\frac {n^2 + n}{2} = 0$, so P(0) is true.
Inductive Hypothesis: Suppose that P(k) is true for some integer $k \ge 0$.
Then $a_{k+1} = (k + 1) + a_k$
= $(k + 1) + \frac {k^2 + k}{2}$
= $\frac{2(k + 1) + (k^2 + k)}{2}$
= $\frac{2k + 2 + k^2 + k}{2}$
= $\frac{k^2 + 3k + 2}{2}$
= $\frac{k^2 + 3k}{2} + 1$??? I'm not sure what to do after the previous step. Any help would be greatly appreciated.

Induction isn't actually needed here. Notice that each term is a partial sum, in particular:
$$a_n = \sum_{k=0}^n k.$$ 
You guessed that $a_n = \frac{n^2 +n}{2}$, which is correct, so you can just show that
$$\sum_{k=0}^n k = \frac{n^2 +n}{2}.$$
The usual way is to add the sum here to itself but write it backwards once, i.e., 
\begin{align}2 \sum_{k=0}^n k & = \hspace{.05in} \sum_{k=0}^n k +  \sum_{k=0}^n k\\
& =\hspace{.2in}1 +\hspace{.2in} 2 \hspace{.2in} + ...+(n-1) + n \\
& \hspace{.2in}+ n + (n-1) +... + \hspace{.2in}2 \hspace{.2in}+ 1 \\
& =\hspace{.05in} n(n+1)\\
& =\hspace{.05in} n^2 +n .
\end{align}
In the second to last step we added vertically to get $n+1$, $n$ times. Then just divide by two.
