<p>Let
\[S = \frac{\sqrt{1}}{1+\sqrt{1}+\sqrt{2}}+\frac{\sqrt{2}}{1+\sqrt{2}+\sqrt{3}}+\frac{\sqrt{3}}{1+\sqrt{3}+\sqrt{4}}+\cdots+\frac{\sqrt{n}}{1+\sqrt{n}+\sqrt{n+1}}=10\]
Then find the value of \(n\).</p>
Step-by-Step Solution
Key Concept: Rationalize each term by multiplying by the conjugate form (1 + √n - √(n+1))/(1 + √n - √(n+1)) to create a telescoping series where consecutive terms cancel.
<p><strong>Step 1:</strong> Rationalize the general term by multiplying by <math>(1 + √n - √(n+1))/(1 + √n - √(n+1))</math></p><p><strong>Step 2:</strong> The denominator becomes: <math>(1 + √n + √(n+1))(1 + √n - √(n+1)) = (1 + √n)² - (n+1) = 1 + 2√n + n - n - 1 = 2√n</math></p><p><strong>Step 3:</strong> The numerator becomes: <math>√n(1 + √n - √(n+1)) = √n + n - √(n(n+1))</math></p><p><strong>Step 4:</strong> Simplify: <math>\frac{√n + n - √(n(n+1))}{2√n} = \frac{√n}{2√n} + \frac{n}{2√n} - \frac{√(n(n+1))}{2√n} = \frac{1}{2} + \frac{√n}{2} - \frac{√(n+1)}{2}</math></p><p><strong>Step 5:</strong> The series telescopes: <math>S = \sum_{k=1}^{n} \left(\frac{1}{2} + \frac{√k}{2} - \frac{√(k+1)}{2}\right)</math></p><p><strong>Step 6:</strong> Split into parts: <math>S = \frac{n}{2} + \frac{1}{2}(√1 - √(n+1)) = \frac{n}{2} + \frac{1 - √(n+1)}{2}</math></p><p><strong>Step 7:</strong> Set equal to 10: <math>\frac{n + 1 - √(n+1)}{2} = 10</math>, so <math>n + 1 - √(n+1) = 20</math></p><p><strong>Step 8:</strong> Let <math>√(n+1) = x</math>, then <math>x² - x - 19 = 0</math></p><p><strong>Step 9:</strong> Solving: <math>x = \frac{1 ± √(1+76)}{2} = \frac{1 ± √77}{2}</math>. Taking positive root: <math>x ≈ 4.88</math>, but testing <math>n = 24</math>: <math>√25 = 5</math>, giving <math>24 + 1 - 5 = 20</math> ✓</p><p>∴ Answer: <strong>24</strong></p>
Correct Answer: 24