Prove the following by using the principle of mathematical induction for all $n \in N:$
$a+(a+d)+(a+2d)+\ldots+(a+(n-1)d) = \frac{n}{2}[2a+(n-1)d]$

(A) Let $P(n)$ be the statement: $a+(a+d)+(a+2d)+\ldots+(a+(n-1)d) = \frac{n}{2}[2a+(n-1)d]$.
Step $1$: For $n=1$,the $LHS$ is $a$ and the $RHS$ is $\frac{1}{2}[2a+(1-1)d] = \frac{1}{2}(2a) = a$. Since $LHS$ = $RHS$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$a+(a+d)+\ldots+(a+(k-1)d) = \frac{k}{2}[2a+(k-1)d]$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$a+(a+d)+\ldots+(a+(k-1)d) + (a+kd) = \frac{k+1}{2}[2a+kd]$.
Starting from the $LHS$ of $P(k+1)$:
$= \frac{k}{2}[2a+(k-1)d] + (a+kd)$
$= \frac{2ak + k(k-1)d + 2a + 2kd}{2}$
$= \frac{2a(k+1) + (k^2-k+2k)d}{2}$
$= \frac{2a(k+1) + (k^2+k)d}{2}$
$= \frac{2a(k+1) + k(k+1)d}{2}$
$= \frac{k+1}{2}[2a+kd]$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true. By the principle of mathematical induction,$P(n)$ is true for all $n \in N$.