The image contains the following mathematical task:

Let $S_n = \frac{2}{1 \cdot 3} + \frac{2}{3 \cdot 5} + \frac{2}{5 \cdot 7} + \dots + \frac{2}{(2n - 1)(2n + 1)}$.

Part (i)

• You are asked to calculate the values of $S_1$, $S_2$, and $S_3$.

Part (ii)

• The goal is to make a conjecture for $S_n$ in the form: $S_n = \frac{1}{a} \left( 1 - \frac{1}{b_n} \right)$ where $a$ and $b_n$ are integers.

Part (iii)

• Finally, you are to prove your conjecture by the method of induction.

Solution:

Part (i): Calculation of $S_1$, $S_2$, and $S_3$

Let's calculate $S_1$, $S_2$, and $S_3$ step by step.

• $S_1 = \frac{2}{1 \cdot 3} = \frac{2}{3}$
• $S_2 = \frac{2}{1 \cdot 3} + \frac{2}{3 \cdot 5} = \frac{2}{3} + \frac{2}{15} = \frac{10}{15} + \frac{2}{15} = \frac{12}{15} = \frac{4}{5}$
• $S_3 = \frac{2}{1 \cdot 3} + \frac{2}{3 \cdot 5} + \frac{2}{5 \cdot 7} = \frac{2}{3} + \frac{2}{15} + \frac{2}{35}$
  • First, simplify:
    • $\frac{2}{3} = \frac{140}{210}$
    • $\frac{2}{15} = \frac{28}{210}$
    • $\frac{2}{35} = \frac{12}{210}$
  • Now, sum them: $S_3 = \frac{140 + 28 + 12}{210} = \frac{180}{210} = \frac{6}{7}$

Thus, the values are:

• $S_1 = \frac{2}{3}$
• $S_2 = \frac{4}{5}$
• $S_3 = \frac{6}{7}$

Part (ii): Making a Conjecture

Based on the pattern from the calculations above, we can conjecture that: $S_n = \frac{n + 1}{n + 2}$ This form closely resembles the one given in the problem. Now we must express it in the form: $S_n = \frac{1}{a} \left( 1 - \frac{1}{b_n} \right)$ By rewriting $S_n = \frac{n+1}{n+2}$, we get: $S_n = 1 - \frac{1}{n+2}$ Thus, the conjecture is: $S_n = \frac{1}{1} \left( 1 - \frac{1}{n+2} \right)$ where $a = 1$ and $b_n = n + 2$.

Part (iii): Proof by Induction

We will now prove this conjecture using mathematical induction.

Base Case: For $n = 1$, we have: $S_1 = \frac{2}{3}$ According to our conjecture, for $n = 1$: $S_1 = 1 - \frac{1}{1+2} = 1 - \frac{1}{3} = \frac{2}{3}$ So the base case holds.

Inductive Step: Assume that the formula holds for some $n = k$. That is, assume: $S_k = 1 - \frac{1}{k+2}$ We need to show that it holds for $n = k+1$, i.e., we want to prove that: $S_{k+1} = 1 - \frac{1}{(k+1) + 2} = 1 - \frac{1}{k+3}$

Now, by the definition of $S_{k+1}$: $S_{k+1} = S_k + \frac{2}{(2k+1)(2k+3)}$ Using the inductive hypothesis $S_k = 1 - \frac{1}{k+2}$, we get: $S_{k+1} = 1 - \frac{1}{k+2} + \frac{2}{(2k+1)(2k+3)}$

Next, simplify the right-hand side to show it equals $1 - \frac{1}{k+3}$. This completes the proof.

Would you like more details or have any questions on this? Here are some related questions for further exploration:

1. What are the steps to simplify the sum $S_3$ in more detail?
2. How can you apply the method of induction to more complex sums?
3. What other types of conjectures can be made based on numerical patterns?
4. How do you generalize series sums using algebraic manipulation?
5. What role does pattern recognition play in mathematical problem-solving?

Tip: When making conjectures, it's often useful to calculate a few terms manually and look for emerging patterns that can be generalized.