Question

If \(P = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30}\), then find the value of \(\sqrt{6(9-P)}\).

• A. 5
• B. 6
• C. 7
• D. 8

Hint:

\(\frac{1}{n(n+1)} = \frac{1}{n}-\frac{1}{n+1}\)

Answer 1

The Correct Option is C

Step 1: Write each term as a difference of reciprocals. 
\(\frac{1}{2} = 1 - \frac{1}{2}\) 
\(\frac{1}{6} = \frac{1}{2} - \frac{1}{3}\) 
\(\frac{1}{12} = \frac{1}{3} - \frac{1}{4}\) 
\(\frac{1}{20} = \frac{1}{4} - \frac{1}{5}\) 
\(\frac{1}{30} = \frac{1}{5} - \frac{1}{6}\) 

Step 2: Sum the series (telescoping). 
\(P = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{6})\). 
All intermediate terms cancel. 
\(P = 1 - \frac{1}{6} = \frac{5}{6}\). 

Step 3: Compute \(9 - P\). 
\(9 - \frac{5}{6} = \frac{54}{6} - \frac{5}{6} = \frac{49}{6}\). 

Step 4: Compute \(6(9-P)\) and its square root. 
\(6 \times \frac{49}{6} = 49\). 
\(\sqrt{49} = 7\).