Arithmetic Series Proof

Question

Prove : $(a+d)+(a+2d)+(a+3d)+...+(a+nd)=n[\frac{(a+d)+(a+nd)}{2}]$

I have tried proving this by induction but I keep getting stuck at the k+1 step. I have : $k[\frac{(a+d)+(a+kd)}{2}]+(a+(k+1)d)$ which I cannot get to simplify properly.

Perhaps I am doing something wrong or induction is not the proper approach.

Answer 1

Continuing from what you are doing, $$k\left[\frac{(a+d)+(a+kd)}{2}\right]+\left(a+(k+1)\cdot d\right)\\=k\cdot a + \frac{k(k+1)}{2}\cdot d+a+(k+1)\cdot d\\=(k+1)\cdot a+\frac{(k+1)(k+2)}{2}\cdot d\\ =(k+1)\cdot\left[\frac{2a+(k+2)\cdot d}{2}\right]\\=(k+1)\cdot\left[\frac{(a+d)+(a+(k+1)\cdot d)}{2}\right]\\$$

Answer 2

Hint: $2(a+(k+1)d) = 2a +2kd + 2d = (a+d) + (a+kd) + (k+1)d$