$T(n) = T(n-1)+n$
$=T(n-2)+(n-1)+n$
$=T(n-3)+(n-2)+(n-1)+n$
.
.
.
$=T(n-k)+\left[(n-k+1)+ (n-k+2)+\ldots+(n-1) + n\right]$
Recurrence stops at,
$n-k=1$
$k=n-1$
$\therefore T(n) = T(1)+\left[2+3+\ldots+n\right]
\\= 1+ 2+3+\ldots + n
\\= \frac{n(n+1)}{2}$
PS: Unless explicitly asked for asymptotic bound, we should always try to get the exact answer.