Turning a recursively defined sequence into an explicit formula 
I have been given the following recursive sequence:
  $$a_0=0,\quad\text{and}\quad a_k=k+a_k-1\ \text{for each integer $k\ge 1$}$$
  and I need to give the explicit formula. 

I have written out the sequence and can recognize the pattern but can't figure out how to write an explicit formula so any help would be appreciated.
Thanks in advance :)
 A: Use generating functions. Define $A(z) = \sum_{k \ge 0} a_k z^k$, shift your recurrence by one, multiply by $z^k$ and sum over $k \ge 0$, recognize sums:
$\begin{align*}
  \sum_{k \ge 0} a_{k + 1} z^k
    &= \sum_{n \ge 0} (k + 1) z^k + \sum_{k \ge 0} a_k z^k \\
  \frac{A(z) - a_0}{z}
    &= \frac{d}{d z} \frac{1}{1 - z} + A(z) \\
    &= \frac{1}{(1 - z)^2} + A(z)
\end{align*}$
Solve for $A(z)$, extract coefficients:
$\begin{align*}
    A(z)
      &= \frac{z}{(1 - z)^3} \\
    a_k
      &= [z^k] A(z) \\
      &= [z^{k - 1}] (1 - z)^{-3} \\
      &= (-1)^{k - 1} \binom{-3}{k - 1} \\
      &= \binom{k - 1 + 3 - 1}{3 - 1} \\
      &= \frac{k (k + 1)}{2}
\end{align*}$
A: Use the method of finite differences. I've worked out this example (Why does the method of undetermined coefficients always work?). 
You just need to work out the recursive sequence for just the first let's say 5 consecutive numbers. And take the difference from them and  go on untill you have a constant row. 
A: I suppose that you mean $$a_k=k+a_{k-1}\tag 1$$ Since you have $k$ in the rhs, "admit" that the solution will be a quadratic expression in $k$, say $$a_k=A+B k +C k^2\tag 2$$ Replace in $(1)$, group terms to get $$(B-C)+(2 C-1) k=0\tag 3$$ Since this has to be true for all $k$, then $B-C=0$ and $2C-1=0$ that is to say $B=C=\frac12$. So
$$a_k=A+\frac{1}{2} k (k+1)\tag4$$ and, since $a_0=0$, $A=0$ and $$a_k=\frac{1}{2} k (k+1)\tag5$$
A: From the theory of linear recurrences, you (should) know that the solution is the combination of a general solution of the homogeneous equation and a particular solution of the non-homogeneous equation.
The first contribution is that of $a_k=a_{k-1}$, obviously an arbitrary constant.
The second is any solution of $a_k-a_{k-1}=k$, where the RHS is a linear polynomial in $k$. By educated guess, a solution could be a quadratic polynomial, let $pk^2+qk+r$.
Then
$$a_k-a_{k-1}=p(2k+1)+q=k$$ and any vaues of $p,q,r$ that satisfy this identity can do.
