How I can solve $s(n)=n+s(n-1)$ by iteration method? $$
s(n)=
\begin{cases}
0, \text{if $n=0$}\\
n+s(n-1), \text{if $n>0$}\\
\end{cases}
$$
Using the relation
\begin{align}
s(n) &= n+s(n-1) \\
&= n+n-1+s(n-2) \\
&= n+n-1+n-2+s(n-3) \\
&= \dots \\
&= n+n-1+n-2+\dots+1 \\
&= \dfrac{n^2+n}2 \\
&= \theta(n^2)
\end{align}
Is it right?

I like to turn
recurrences to
telescoping sums
whenever possible.
From
$s(n) = n+s(n-1)
$
I get
$s(n)-s(n-1) = n$.
Summing both sides from $1$ to $m$,
$\sum_{n=1}^m s(n)-s(n-1) = \sum_{n=1}^m n $.
The left side telescopes into
$s(m)-s(0)$
and the right side is just
$\dfrac{m(m+1)}{2}
$.
Note that if
the first value is $s(1)$,
just make the sum from $2$
instead of $1$.
