Let’s address each question from the document systematically:

---

1. Proof by Mathematical Induction for $P(n)$:

a. What is the statement $P(1)$?

$P(1): 1^2 = \frac{1(1+1)(2(1)+1)}{6}$

b. Show that $P(1)$ is true.

• Left-hand side: $1^2 = 1$
• Right-hand side: $\frac{1(1+1)(2(1)+1)}{6} = \frac{1(2)(3)}{6} = 1$
• Since $LHS = RHS$, $P(1)$ is true.

c. What is the inductive hypothesis?

Assume $P(k)$ is true: $1^2 + 2^2 + \dots + k^2 = \frac{k(k+1)(2k+1)}{6}$

d. What do you need to prove in the inductive step?

Prove $P(k+1)$: $1^2 + 2^2 + \dots + k^2 + (k+1)^2 = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$

e. Complete the inductive step:

1. Start from the hypothesis: $1^2 + 2^2 + \dots + k^2 = \frac{k(k+1)(2k+1)}{6}$
2. Add $(k+1)^2$ to both sides: $1^2 + 2^2 + \dots + k^2 + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2$
3. Simplify the RHS: $= \frac{k(k+1)(2k+1) + 6(k+1)^2}{6}$ $= \frac{(k+1)\big[k(2k+1) + 6(k+1)\big]}{6}$ $= \frac{(k+1)\big[2k^2 + k + 6k + 6\big]}{6}$ $= \frac{(k+1)\big[2k^2 + 7k + 6\big]}{6}$ $= \frac{(k+1)(k+2)(2k+3)}{6}$
4. This matches the formula for $P(k+1)$.

Thus, by induction, $P(n)$ is true for all $n \geq 1$.

---

2. Prove the Sum Formula:

Prove: $1 \cdot 2 + 2 \cdot 3 + \dots + n(n+1) = \frac{n(n+1)(n+2)}{3}$

Proof by Induction:

• Base case ($n = 1$): $1 \cdot 2 = \frac{1(1+1)(1+2)}{3} = \frac{1 \cdot 2 \cdot 3}{3} = 2$ $LHS = RHS$, so true for $n = 1$.

• Inductive Hypothesis: Assume true for $n = k$: $1 \cdot 2 + 2 \cdot 3 + \dots + k(k+1) = \frac{k(k+1)(k+2)}{3}$

• Inductive Step ($n = k+1$): Add $(k+1)(k+2)$ to both sides: $1 \cdot 2 + 2 \cdot 3 + \dots + k(k+1) + (k+1)(k+2) = \frac{k(k+1)(k+2)}{3} + (k+1)(k+2)$ Factor $(k+1)(k+2)$ on the RHS: [ = (k+1)(k+2)\big[\frac{k}{3} + 1\big] = (k+1)(k+2)\big[\frac{k+3}{3}\big] ] $= \frac{(k+1)(k+2)(k+3)}{3}$ This matches the formula for $P(k+1)$.

By induction, the formula is true for all $n \geq 1$.

---

3. Evaluate $f(n)$ for Recursively Defined Functions:

a. $f(n+1) = f(n) + 2, f(0) = 1$:

$f(1) = 1 + 2 = 3, f(2) = 3 + 2 = 5, f(3) = 5 + 2 = 7, f(4) = 7 + 2 = 9$

b. $f(n+1) = 3f(n), f(0) = 1$:

$f(1) = 3 \cdot 1 = 3, f(2) = 3 \cdot 3 = 9, f(3) = 3 \cdot 9 = 27, f(4) = 3 \cdot 27 = 81$

c. $f(n+1) = 2f(n), f(0) = 1$:

$f(1) = 2 \cdot 1 = 2, f(2) = 2 \cdot 2 = 4, f(3) = 2 \cdot 4 = 8, f(4) = 2 \cdot 8 = 16$

---

4. Recursive Definitions:

a. $a_n = 6n$:

$a_1 = 6, a_2 = 12, a_3 = 18, a_{n+1} = a_n + 6$

b. $a_n = 10n$:

$a_1 = 10, a_2 = 20, a_3 = 30, a_{n+1} = a_n + 10$

c. $a_n = 2n + 1$:

$a_1 = 3, a_2 = 5, a_3 = 7, a_{n+1} = a_n + 2$

---

5. List Ordered Pairs for Matrices:

a.

Matrix:

1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$$ Ordered pairs: $$(1, 1), (1, 3), (2, 2), (3, 1), (3, 3)$$ #### b. Matrix: $$\begin{bmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \end{bmatrix}$$ Ordered pairs: $$(1, 2), (2, 2), (3, 2)$$ #### c. Matrix: $$\begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix}$$ Ordered pairs: $$(1, 1), (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2), (3, 3)$$ --- If you'd like a continuation with the remaining questions or additional clarifications, feel free to