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Let's solve this by proving each part using the principle of mathematical induction.

### Part (a)
#### Statement:
For all positive integers \( n \):
\[
1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}
\]

#### Proof by Induction:
1. **Base Case** (\( n = 1 \)):
   \[
   1 = \frac{1 \cdot (1 + 1)}{2} = \frac{1 \cdot 2}{2} = 1
   \]
   The base case holds true.

2. **Inductive Step**:
   - Assume the formula holds for some integer \( k \), i.e.,
     \[
     1 + 2 + 3 + \ldots + k = \frac{k(k+1)}{2}
     \]
   - Now, prove that it holds for \( k + 1 \), i.e., show that:
     \[
     1 + 2 + 3 + \ldots + k + (k + 1) = \frac{(k + 1)(k + 2)}{2}
     \]
   - Starting with the inductive hypothesis:
     \[
     1 + 2 + 3 + \ldots + k + (k + 1) = \frac{k(k+1)}{2} + (k + 1)
     \]
   - Factor out \( (k + 1) \) from the right side:
     \[
     = \frac{k(k+1) + 2(k+1)}{2} = \frac{(k+1)(k+2)}{2}
     \]
   - This matches the required expression for \( k + 1 \).

Since both the base case and inductive step are proven, the statement holds for all positive integers \( n \).

---

### Part (b)
#### Statement:
For all positive integers \( n \):
\[
1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + n(n+1) = \frac{n(n+1)(n+2)}{3}
\]

#### Proof by Induction:
1. **Base Case** (\( n = 1 \)):
   \[
   1 \cdot 2 = \frac{1 \cdot 2 \cdot 3}{3} = \frac{6}{3} = 2
   \]
   The base case holds true.

2. **Inductive Step**:
   - Assume the formula holds for some integer \( k \), i.e.,
     \[
     1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + k(k+1) = \frac{k(k+1)(k+2)}{3}
     \]
   - Now, prove that it holds for \( k + 1 \), i.e., show that:
     \[
     1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + k(k+1) + (k + 1)(k + 2) = \frac{(k + 1)(k + 2)(k + 3)}{3}
     \]
   - Starting with the inductive hypothesis:
     \[
     1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + k(k+1) + (k + 1)(k + 2) = \frac{k(k+1)(k+2)}{3} + (k + 1)(k + 2)
     \]
   - Factor out \( (k + 1)(k + 2) \) from the right side:
     \[
     = \frac{k(k+1)(k+2) + 3(k+1)(k+2)}{3} = \frac{(k+1)(k+2)(k+3)}{3}
     \]
   - This matches the required expression for \( k + 1 \).

Since both the base case and inductive step are proven, the statement holds for all positive integers \( n \).

---

Would you like further details or explanations? 

Here are some related questions for practice:
1. What is the principle of mathematical induction?
2. How would you prove a different series sum using induction?
3. Can induction be used to prove inequalities?
4. What are common errors to avoid in inductive proofs?
5. How does induction differ from other proof techniques?

**Tip:** When using induction, always start with a clear base case and ensure each step logically follows from the assumption.

Using the principle of mathematical induction, prove the following statements for all positive integers:

(a) 1 + 2 + 3 + ... + n = n(n + 1) / 2
(b) 1·2 + 2·3 + 3·4 + ... + n(n + 1) = n(n + 1)(n + 2) / 3

Learn how to prove series summation formulas using the principle of mathematical induction. This problem covers the sum of the first n natural numbers and a product sequence, making it suitable for Grades 11-12 students.

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