how should I write the recursion relation using the $\Sigma$ ?
we have $\begin{equation*} T(n) = \begin{cases} 1 &n=0\\ T(n-1)+n &n>0 \end{cases} \end{equation*}$
so if we suppose we reached the end, the recursion relation is : $T(n)=T(n-k)+(n-(k-1))+(n-(k-2))+...+(n-1)+n$
I'm not sure show to do it since the $k$ was decreasing
We have: $$ T(n) = T(n-1) + n + T(n - 2) + (n-1) + ... + T(2 - 1) + 2 + T(1 - 1) + 1 = \sum_{k = 1}^n {k + T(n-k)} $$
Just for fun, I'll write the closed form (it might be helpful for you), we have:
$$ T(k) - T(k - 1) = k\implies \sum_{k=1}^n {T(k) - T(k - 1)} = \sum_{k=1}^n k = \frac{n(n+1)}{2} $$ Now we have a telescoping series: $$ \sum_{k=1}^n T(k) - \sum_{k=1}^n T(k-1) = \sum_{k=1}^n T(k) - \sum_{k=0}^{n-1} T(k) = T(n)-T(0) = \frac{n(n+1)}{2}\implies T(n) = \frac{n(n+1)}{2}+T(0) \implies \boxed{T(n)= \frac{n(n+1)}{2} + 1} $$
