Here is the solution to problem 2 using the Principle of Mathematical Induction. 2. Prove that $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$, $\forall n \ge 1$. Let $P(n)$ be the statement $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$. Base Case (n=1): LHS $= \sum_{k=1}^{1} k^2 = 1^2 = 1$ RHS $= \frac{1(1+1)(2(1)+1)}{6} = \frac{1(2)(3)}{6} = \frac{6}{6} = 1$ Since LHS = RHS, $P(1)$ is true. Inductive Hypothesis: Assume $P(m)$ is true for some positive integer $m$. That is, assume $\sum_{k=1}^{m} k^2 = \frac{m(m+1)(2m+1)}{6}$. Inductive Step: We need to prove that $P(m+1)$ is true, i.e., $\sum_{k=1}^{m+1} k^2 = \frac{(m+1)((m+1)+1)(2(m+1)+1)}{6} = \frac{(m+1)(m+2)(2m+3)}{6}$. Consider the LHS of $P(m+1)$: $$ \sum_{k=1}^{m+1} k^2 = \left( \sum_{k=1}^{m} k^2 \right) + (m+1)^2 $$ By the Inductive Hypothesis, substitute $\sum_{k=1}^{m} k^2 = \frac{m(m+1)(2m+1)}{6}$: $$ \frac{m(m+1)(2m+1)}{6} + (m+1)^2 $$ Factor out $(m+1)$: $$ (m+1) \left[ \frac{m(2m+1)}{6} + (m+1) \right] $$ Find a common denominator: $$ (m+1) \left[ \frac{m(2m+1)}{6} + \frac{6(m+1)}{6} \right] $$ $$ (m+1) \left[ \frac{2m^2+m + 6m+6}{6} \right] $$ $$ (m+1) \left[ \frac{2m^2+7m+6}{6} \right] $$ Factor the quadratic $2m^2+7m+6$: $$ 2m^2+7m+6 = 2m^2+4m+3m+6 = 2m(m+2)+3(m+2) = (2m+3)(m+2) $$ Substitute the factored quadratic back into the expression: $$ (m+1) \frac{(m+2)(2m+3)}{6} $$ $$ \frac{(m+1)(m+2)(2m+3)}{6} $$ This is the RHS of $P(m+1)$. Thus, $P(m+1)$ is true. By the Principle of Mathematical Induction, $P(n)$ is true for all $n \ge 1$. The proof is complete.