Sequences & Series
Telescoping Series
Grade 11
Question:
<p>\(\frac{1}{\sqrt{2}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{11}}+\cdots\) \(n\) terms is equal to</p>
<p>\(\frac{\sqrt{3n+2}-\sqrt{2}}{3}\)</p>
<p>\(\frac{n}{\sqrt{2+3n}+\sqrt{2}}\)</p>
<p>less than \(n\)</p>
<p>less than \(\sqrt{\dfrac{n}{3}}\)</p>
Step-by-Step Solution
Key Concept: Rationalize each term by multiplying by the conjugate to create a telescoping series where consecutive terms cancel, leaving only the first and last terms.
<p><strong>Step 1:</strong> Identify the general term. The denominators follow the pattern: (√2 + √5), (√5 + √8), (√8 + √11), ... where consecutive terms differ by √3. The <em>k</em>-th term has denominator √(3<em>k</em>-1) + √(3<em>k</em>+2).</p><p><strong>Step 2:</strong> Rationalize each term by multiplying numerator and denominator by the conjugate:</p><p>$$\frac{1}{\sqrt{3k-1}+\sqrt{3k+2}} \cdot \frac{\sqrt{3k+2}-\sqrt{3k-1}}{\sqrt{3k+2}-\sqrt{3k-1}} = \frac{\sqrt{3k+2}-\sqrt{3k-1}}{(3k+2)-(3k-1)} = \frac{\sqrt{3k+2}-\sqrt{3k-1}}{3}$$</p><p><strong>Step 3:</strong> Write out the series as a telescoping sum:</p><p>$$\sum_{k=1}^{n} \frac{\sqrt{3k+2}-\sqrt{3k-1}}{3} = \frac{1}{3}[(\sqrt{5}-\sqrt{2})+(\sqrt{8}-\sqrt{5})+(\sqrt{11}-\sqrt{8})+\cdots+(\sqrt{3n+2}-\sqrt{3n-1})]$$</p><p><strong>Step 4:</strong> Most terms cancel (telescoping). Only the last positive term and first negative term remain:</p><p>$$= \frac{1}{3}(\sqrt{3n+2}-\sqrt{2})$$</p><p>∴ Answer: <strong>A</strong> (or whichever option matches $\frac{\sqrt{3n+2}-\sqrt{2}}{3}$)</p>
Correct Answer: A