Answer
(a) $1(1+1)=\frac{2(2-1)(2+1)}{3}$, true.
(b) $\sum_{i=1}^{k-1}i(i+1)=\frac{k(k-1)(k+1)}{3}$
(c) $\sum_{i=1}^{k}i(i+1)=\frac{(k+1)k(k+2)}{3}$
(d) if P(k) is true then P(k+1) will also be true
Work Step by Step
(a) For $n=2$, we have $LHS=1(1+1)=2$ and
$RHS=\frac{2(2-1)(2+1)}{3}=2$, thus P(2) is true.
(b) For $P(k)$, we have $\sum_{i=1}^{k-1}i(i+1)=\frac{k(k-1)(k+1)}{3}$
(c) For $P(k+1)$, we have $\sum_{i=1}^{k}i(i+1)=\frac{(k+1)k(k+2)}{3}$
(d) To prove by mathematical induction that the formula holds for all integers $n\ge2$, we must show that if P(k) is true then P(k+1) will also be true in the inductive step.