Answer
Check answer for detailed proof
Work Step by Step
We wish to prove that the statement :
$S_{n}: \sum_{k=0}^{n-1}(a+k d)=\frac{n}{2}[2 a+(n-1) d]$
is valid for all positive integers $n$.
1. (Anchor step.)
We verify that for $\mathrm{n}=1$ (when we have one term in the sum), the formula holds true
$a=\frac{1}{2} a(2 a+(1-1) d$
The statement is valid for $n=1$.
2. Induction step.
If from the assumption that the formula is valid for i $(\mathrm{n}=\mathrm{i})$ we prove that it will be valid for $\mathrm{n}=\mathrm{i}+1$ terms, the formula will be valid for all integers $n$.
(our "usual" index $k$ has been taken, so we use another one, i).
$(*)$ Assume that $S_{i}: \sum_{k=0}^{i-1}(a+k d)=\frac{i}{2}[2 a+(i-1) d]$ is valid
(is $S_{k+1}: \sum_{k=0}^{i-1+1}(a+k d)=\left(\frac{i+1}{2}\right)[2 a+i d]$ valid? $)$
$s_{i+1}=S_{i}+a_{i+1}$
$\sum_{k=0}^{i-1+1}(a+k d)=\frac{i}{2}[2 a+(i-1) d]+[a+i d] \quad$ as we assumed in $(*)$
$=\frac{2 i a+i(i-1) d+2 a+2 i d}{2}$
$=\frac{2 a(i+1)+i d(i+1)}{2}$
$=\left(\frac{i+1}{2}\right)[2 a+i d]$
It is By mathematical induction, the formula is valid for all positive integer values of $n$
