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?

$s(n)=n+s(n-1)$
$s(n-1)=n-1 + s(n-2)$
$s(n-2)=n-2 + s(n-3)$
$s(n-3)=n-3 + s(n-4)$
.
.
.
$s[n-(n-1)]=n-(n-1) + s(0)$
⇒ $s(n)=n\times n -\frac{(n-1)(n-1+1)}{2}=n^2-\frac{(n)(n-1)}{2}=\frac{n^2+n}{2}$
