Answer
$\displaystyle \sum_{k=1}^{n+1}(a+(k-1)d)$
or $\displaystyle \sum_{k=0}^{n}(a+kd)$
Work Step by Step
The first term (when k=$1$) is $a=a+0d=a+(1-1)d$,
The second term (when k=$2$) is $a+d=a+(2-1)d$,
The third term (when k=$3$) is $a+2d=a+(3-1)d$,
The kth is $a+(k-1)d$.
We must have n+1 terms, as the last must be $a+(n+1-1)d=a+nd$
$a+(a+d)+(a+2d)+\displaystyle \dots+(a+nd)=\sum_{k=1}^{n+1}(a+(k-1)d)$
or, if we count the terms with index k starting from 0
$=\displaystyle \sum_{k=0}^{n}(a+kd)$
