Here are the proofs using the Principle of Mathematical Induction. 3. Prove that $\sum_{k=1}^{n} (2k-1)^2 = \frac{1}{3}n(4n^2-1)$, $\forall n \ge 1$. Step 1: Base Case (n=1) For $n=1$, the LHS is $\sum_{k=1}^{1} (2k-1)^2 = (2(1)-1)^2 = 1^2 = 1$. The RHS is $\frac{1}{3}(1)(4(1)^2-1) = \frac{1}{3}(1)(4-1) = \frac{1}{3}(3) = 1$. Since LHS = RHS, the statement is true for $n=1$. Step 2: Inductive Hypothesis Assume that the statement is true for some integer $k \ge 1$. That is, assume $\sum_{i=1}^{k} (2i-1)^2 = \frac{1}{3}k(4k^2-1)$. Step 3: Inductive Step (n=k+1) We need to prove that the statement is true for $n=k+1$: $$ \sum_{i=1}^{k+1} (2i-1)^2 = \frac{1}{3}(k+1)(4(k+1)^2-1) $$ Consider the LHS: $$ \sum_{i=1}^{k+1} (2i-1)^2 = \left(\sum_{i=1}^{k} (2i-1)^2\right) + (2(k+1)-1)^2 $$ Using the inductive hypothesis, we substitute $\sum_{i=1}^{k} (2i-1)^2 = \frac{1}{3}k(4k^2-1)$: $$ = \frac{1}{3}k(4k^2-1) + (2k+2-1)^2 $$ $$ = \frac{1}{3}k(4k^2-1) + (2k+1)^2 $$ Factor $4k^2-1$ as $(2k-1)(2k+1)$: $$ = \frac{1}{3}k(2k-1)(2k+1) + (2k+1)^2 $$ Factor out $(2k+1)$: $$ = (2k+1)\left[\frac{k(2k-1)}{3} + (2k+1)\right] $$ $$ = (2k+1)\left[\frac{2k^2-k + 3(2k+1)}{3}\right] $$ $$ = (2k+1)\left[\frac{2k^2-k+6k+3}{3}\right] $$ $$ = (2k+1)\left[\frac{2k^2+5k+3}{3}\right] $$ Factor the quadratic $2k^2+5k+3$: $$ 2k^2+5k+3 = 2k^2+2k+3k+3 = 2k(k+1)+3(k+1) = (2k+3)(k+1) $$ Substitute this back into the expression: $$ = (2k+1)\frac{(k+1)(2k+3)}{3} $$ $$ = \frac{1}{3}(k+1)(2k+1)(2k+3) $$ Now, let's expand the RHS we want to achieve: $$ \frac{1}{3}(k+1)(4(k+1)^2-1) = \frac{1}{3}(k+1)(4(k^2+2k+1)-1) $$ $$ = \frac{1}{3}(k+1)(4k^2+8k+4-1) $$ $$ = \frac{1}{3}(k+1)(4k^2+8k+3) $$ Factor the quadratic $4k^2+8k+3$: $$ 4k^2+8k+3 = 4k^2+2k+6k+3 = 2k(2k+1)+3(2k+1) = (2k+1)(2k+3) $$ So, the RHS is: $$ = \frac{1}{3}(k+1)(2k+1)(2k+3) $$ Since LHS = RHS, the statement is true for $n=k+1$. By the Principle of Mathematical Induction, the statement $\sum_{k=1}^{n} (2k-1)^2 = \frac{1}{3}n(4n^2-1)$ is true for all integers $n \ge 1$. 4. Prove that $n^3 - n$ is divisible by 6 for all $n \ge 1$. Step 1: Base Case (n=1) For $n=1$, $1^3 - 1 = 1 - 1 = 0$. Since $0$ is divisible by $6$ ($0 = 6 \times 0$), the statement is true for $n=1$. Step 2: Inductive Hypothesis Assume that the statement is true for some integer $k \ge 1$. That is, assume $k^3 - k$ is divisible by 6. This means $k^3 - k = 6m$ for some integer $m$. Step 3: Inductive Step (n=k+1) We need to prove that $(k+1)^3 - (k+1)$ is divisible by 6. Consider the expression: $$ (k+1)^3 - (k+1) $$ Expand $(k+1)^3$: $$ = (k^3 + 3k^2 + 3k + 1) - (k+1) $$ $$ = k^3 + 3k^2 + 3k + 1 - k - 1 $$ Rearrange the terms to use the inductive hypothesis: $$ = (k^3 - k) + 3k^2 + 3k $$ $$ = (k^3 - k) + 3k(k+1) $$ By the inductive hypothesis, $k^3 - k$ is divisible by 6. So, we can write $k^3 - k = 6m$ for some integer $m$. Now consider the term $3k(k+1)$. The product of two consecutive integers, $k(k+1)$, is always an even number. This is because either $k$ is even or $k+1$ is even. Therefore, $k(k+1) = 2p$ for some integer $p$. Substitute this into $3k(k+1)$: $$ 3k(k+1) = 3(2p) = 6p $$ So, $3k(k+1)$ is also divisible by 6. Now substitute these back into the expression for $(k+1)^3 - (k+1)$: $$ (k+1)^3 - (k+1) = (k^3 - k) + 3k(k+1) $$ $$ = 6m + 6p $$ $$ = 6(m+p) $$ Since $m$ and $p$ are integers, their sum $m+p$ is also an integer. Thus, $(k+1)^3 - (k+1)$ is a multiple of 6, which means it is divisible by 6. By the Principle of Mathematical Induction, the statement $n^3 - n$ is divisible by 6 is true for all integers $n \ge 1$.