Prove the following by the principle of mathematical induction: `a+(a+d)+(a+2d)++(a+(n-1)d)=n/2[2a+(n-1)d]`

Explanation

Let `n=1`

Then

`a+(a+d)+(a+2d)+ ... +[a+(n-1)d] = a`


On the other hand,
`n/2[2a+(n-1)d] = 1/2 * 2a = a`


So these expressions
...