Question

Find the value of $(1 - \frac{1}{2})(1 - \frac{1}{3})(1 - \frac{1}{4}) \dots (1 - \frac{1}{n})$.

• A. $\frac{1}{n}$
• B. $\frac{1}{n+1}$
• C. $1 + \frac{1}{n}$
• D. $1 - \frac{1}{n}$

Hint:

For products or sums of series, always simplify the first few terms to identify a telescoping pattern where intermediate terms cancel out. This is a common technique for solving such problems.

Answer 1

The Correct Option is A

Step 1: Define \( P(k) = \left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right) \cdots \left(1-\frac{1}{k}\right) \) and check the smallest case, \( k=2 \): \( P(2) = 1-\frac{1}{2} = \frac{1}{2} \), which equals \( \frac{1}{k} \) for \( k=2 \).

Step 2: Assume the pattern \( P(k) = \frac{1}{k} \) holds for some \( k \), and check the next step: \( P(k+1) = P(k) \times \left(1-\frac{1}{k+1}\right) = \frac{1}{k} \times \frac{k}{k+1} = \frac{1}{k+1} \).

Step 3: Since the formula \( \frac{1}{k} \) reproduces itself correctly at each new step, it holds for every \( k \), including \( k=n \), so \( P(n) = \frac{1}{n} \).
\[ \boxed{\frac{1}{n}} \]