Prove that sum_k=1ⁿ (2k−1)² = (1)/(3)n(4n²-1), n 1.

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 $n \ge 1$.