Show $a+(a+d)+(a+2d)+\cdots+(a+nd)=a(n+1)+d\frac{n(n+1)}{2}$ 
Show $a+(a+d)+(a+2d)+\cdots+(a+nd)=a(n+1)+d\frac{n(n+1)}{2}$, where $a$ and $d$ are real numbers and $n$ is an integer.

Attempt:
I first added twice
$$a+(a+d)+(a+2d)+\cdots+(a+nd)$$
to itself in the following way:
$$[a+(a+d)+(a+2d)+\cdots+(a+nd)] + [(a+nd)+(a+(n-1)d)+(a+(n-2)d)+\cdots+a]$$
which equals $(2a+nd)+(2a+nd)+\cdots+(2a+nd)$ which are n quantities.
These $n$ quantities can then be combined to
$$2[a+(a+d)+(a+2d)+\cdots+(a+nd)]=n(2a+nd)$$
$[a+(a+d)+(a+2d)+\cdots+(a+nd)]=\frac{n(2a+nd)}{2}$ divide by $2$
After that point, I begin to screw up and don't find myself any closer to 
$$a(n+1)+d\frac{n(n+1)}{2}$$

Given $$ S = a+(a+d)+(a+2d)+\cdots+(a+nd) \tag 1$$
writting in reverse order,we get
$$ S = (a+nd)+(a+(n-1)d)+\cdots+a \tag 2$$
So adding these two, we get
$$2S = \left[(2a+nd)+(2a+nd)+\cdots+(2a+nd)\right] = (n+1)\cdot (2a+nd)$$
So we get $$S = \frac{(n+1)}{2}\left[2a+nd\right]$$
