Question

The value of \( \sum_{k=1}^{n} \frac{1}{\sqrt{a_{k}} + \sqrt{a_{k+1}}} \), where \( a_{1}, a_{2}, \dots, a_{n} \) are in A.P. with common difference \( d \), is:

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

Hint:

Rationalization often turns a complex sum into a simple telescoping series.

Answer 1

The Correct Option is A

Step 1: Concept
Rationalize each term in the summation.
Step 2: Analysis
Term $= \frac{\sqrt{a_{k+1}} - \sqrt{a_{k}}}{a_{k+1} - a_{k}} = \frac{\sqrt{a_{k+1}} - \sqrt{a_{k}}}{d}$.
The sum is telescoping: $\frac{1}{d} [(\sqrt{a_{2}} - \sqrt{a_{1}}) + (\sqrt{a_{3}} - \sqrt{a_{2}}) + \dots + (\sqrt{a_{n+1}} - \sqrt{a_{n}})]$.
Sum $= \frac{\sqrt{a_{n+1}} - \sqrt{a_{1}}}{d}$.
Step 3: Conclusion
Substitute $d = \frac{a_{n+1} - a_{1}}{n}$. Sum $= \frac{\sqrt{a_{n+1}} - \sqrt{a_{1}}}{(a_{n+1} - a_{1})/n} = \frac{n}{\sqrt{a_{n+1}} + \sqrt{a_{1}}}$.
Final Answer: (A)