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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.

bob
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  • I would have done it this way: $$ \begin{align} (a+d)+(a+2d)+...+(a+nd) &= a+a+...+a + (1+2+...+n)d \ &= na + \frac{n(n+1)}{2}d \ &= n \frac{a+d+a+nd}{2} \end{align} $$ But induction works as well – Random Tourist Apr 06 '17 at 13:55

2 Answers2

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

gandalf61
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Continuing from what you are doing, $$k\left[\frac{(a+d)+(a+kd)}{2}\right]+\left(a+(k+1)\cdot d\right)\\=k\cdot a + \frac{k(k+1)}{2}\cdot d+a+(k+1)\cdot d\\=(k+1)\cdot a+\frac{(k+1)(k+2)}{2}\cdot d\\ =(k+1)\cdot\left[\frac{2a+(k+2)\cdot d}{2}\right]\\=(k+1)\cdot\left[\frac{(a+d)+(a+(k+1)\cdot d)}{2}\right]\\$$

Lazy Lee
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