Sum of $n$ advanced

$a+(a+1)+(a+2)+(a+3)+(a+4)+(a+5)+\dots+(a+n)$

$= (n + 1)a + (1 + 2 + 3 + 4 + 5 +\dots+n)$

$= (n + 1)a + \frac{n(n + 1)}{2}$

$= \frac{2a(n + 1)}{2} + \frac{n(n + 1)}{2}$

$= \frac{(2a + n)(n + 1)}{2}$

$= \frac{(n + 1)(2a + n)}{2}$

$\displaystyle\sum_{k=a}^{n+a} k = \displaystyle\sum_{j=0}^{n}(a+j) =\displaystyle\sum_{j=0}^{n} a + \displaystyle\sum_{j=0}^{n} j$
$\displaystyle\sum_{j=0}^{n} a + \displaystyle\sum_{j=0}^{n} j = a \displaystyle\sum_{j=0}^{n} 1 + \displaystyle\sum_{j=1}^{n} j$ since a is considered to be constant over the series
$\displaystyle\sum_{j=0}^{n} 1 = (n+1)$ (adding one n+1 times) and $\displaystyle\sum_{j=1}^{n} j = \frac{n(n+1)}{2}$ by gauss's formula.
$a \displaystyle\sum_{j=0}^{n} 1 + \displaystyle\sum_{j=1}^{n} j = a(n+1) + \frac{1}{2}n(n+1) = (n+1)(a+\frac{n}{2})$ by factoring
$(n+1)(a+\frac{n}{2}) = \frac{1}{2}(2a+n)(n+1)$ which is equivalent to the first option

Since $\sum_{k=a}^{n+k} k = \sum_{k=1}^{n+k} k - \sum_{k=1}^{a-1} k$ it equals $\frac{\left(a+n\right)\left(a+n+1\right)}{2}-\frac{\left(a-1\right)a}{2} = \frac{1}{2}\left(2a + 2an + n^2 + n\right) = \frac{1}{2}\left(n+1\right)\left(n+2a\right)$

I just considered the case a=0 and found that the 3rd one is correct.

$a+(a+1)+(a+2)+(a+3)+(a+4)+(a+5)+\dots+(a+n)=a(n+1)+\frac{n(n+1)}{2}=\frac{(2a+n)(n+1)}{2}$