Note that starting with your first element, $a_0=1$; to get to $a_1=2$ we sum $1$, to get to $a_2=4$, we sum $2$, to get to $a_3=7$, we sum $3$. So in general, we can say that
$a_{n}=a_{n-1}+n$
This is
$ \begin{align} 2&=1+1 \\[8pt] 4& =2+2\\[8pt] 7&=4+3\\[8pt] 11& =7+4\\[8pt] \cdots&=\cdots\\[8pt] a_n&=a_{n-1}+n \end{align} $
We can get several solutions to you problem, I will share $2$:
Solution 1 Use the recursion to obtain a closed formula:
Since we know that $a_{n}=a_{n-1}+n$ we can write
$a_{n-1}=a_{n-2}+n-1$
so we get $a_{n}=a_{n-2}+(n-1)+n$
Repeating this process, we get that
$a_{n}=a_{n-3}+(n-2)+(n-1)+n$
$a_{n}=a_{n-4}+(n-3)+(n-2)+(n-1)+n$
$a_{n}=a_{n-5}+(n-4)+(n-3)+(n-2)+(n-1)+n$
...so in general we can say that
$a_n=a_{n-k}+(n-k+1)+\cdots+n$ for any $k$ a natural number.
(actually we should be proving the above by induction, but let it be)
So we can choose $k=n$, which means...
$a_n=a_{n-n}+(n-n+1)+\cdots+n$
$a_n=a_0+(1+\cdots+n)$
$a_n=1+\frac{n(n+1)}{2}$
Solution 2
Starting from $a_{n}=a_{n-1}+n$ we use generating functions:
$\eqalign{ & \sum\limits_{n = 1}^\infty {{a_n}} {x^n} = \sum\limits_{n = 1}^\infty {{a_{n - 1}}} {x^n} + \sum\limits_{n = 1}^\infty n {x^n} \cr & \sum\limits_{n = 1}^\infty {{a_n}} {x^n} = x\sum\limits_{n = 1}^\infty {{a_{n - 1}}} {x^{n - 1}} + x\sum\limits_{n = 1}^\infty {n{x^{n - 1}}} \cr & \sum\limits_{n = 0}^\infty {{a_n}} {x^n} - {a_0} = x\sum\limits_{n = 0}^\infty {{a_n}} {x^n} + x\frac{d}{{dx}}\sum\limits_{n = 0}^\infty {{x^n}} \cr & A\left( x \right) - 1 = xA\left( x \right) + x\frac{d}{{dx}}\frac{1}{{1 - x}} \cr & A\left( x \right) - xA\left( x \right) = 1 + \frac{x}{{{{\left( {1 - x} \right)}^2}}} \cr & \left( {1 - x} \right)A\left( x \right) = 1 + \frac{x}{{{{\left( {1 - x} \right)}^2}}} \cr & A\left( x \right) = \frac{1}{{1 - x}} + \frac{x}{{{{\left( {1 - x} \right)}^3}}} \cr} $
The generating sequence of $f(x)=\frac{1}{{1 - x}}$ is $a_n=1$, while the generating sequence of $g(x)=\frac{x}{{{{\left( {1 - x} \right)}^3}}}$ can be obtained by diferentiation of the first one:
$\eqalign{ & \sum\limits_{n = 0}^\infty {{x^n}} = \frac{1}{{1 - x}} \cr & \sum\limits_{n = 1}^\infty {n{x^{n - 1}}} = \frac{1}{{{{\left( {1 - x} \right)}^2}}} \cr & \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){x^{n - 2}}} = \frac{2}{{{{\left( {1 - x} \right)}^3}}} \cr & \sum\limits_{n = 2}^\infty {\frac{{n\left( {n - 1} \right)}}{2}{x^{n - 1}}} = \frac{x}{{{{\left( {1 - x} \right)}^3}}} \cr & \sum\limits_{n = 1}^\infty {\frac{{n\left( {n + 1} \right)}}{2}{x^n}} = \frac{x}{{{{\left( {1 - x} \right)}^3}}} \cr} $
so we finally get that
$a_n=1+{\frac{{n\left( {n + 1} \right)}}{2}}$
ADD I forgot to add the sum of the $a_n$s!
You need to evaluate
$\sum_{k=0}^{n-1} a_k=\sum_{k=0}^{n-1} 1+\sum_{k=0}^{n-1}\frac{k(k+1)}{2}$
$\sum_{k=0}^{n-1} a_k=n+\sum_{k=0}^{n}\frac{k(k-1)}{2}$ $\sum_{k=0}^{n-1} a_k=n+\sum_{k=0}^{n}{k\choose 2}$
Using the binomial identity
$\sum_{k=0}^n {k\choose l}={{n+1}\choose {l+1}}$ we get
(you can find how to obtain it here
$\sum_{k=0}^{n-1} a_k=n+{n+1\choose 3}$
$\sum_{k=0}^{n-1} a_k=n+\frac{(n+1)n(n-1)}{6}$
which is what you wanted.