<p>If \(T_r = \sqrt{r}\sqrt{r+1}\left(\dfrac{4r+5}{(r+2)+\sqrt{r^2-1}}\right)\), then find the value of \(\dfrac{1}{\sqrt{68}}\sum_{r=1}^{16} T_r\).</p>
Step-by-Step Solution
Key Concept: Rationalize the denominator by multiplying with the conjugate, then decompose Tr into a telescoping series where consecutive terms cancel. This transforms a complex sum into a simple difference of first and last terms.
<p><strong>Step 1: Rationalize the denominator</strong></p><p>Multiply by the conjugate: (r+2) - √(r²-1)</p><p>Numerator: (4r+5)[(r+2)-√(r²-1)]</p><p>Denominator: [(r+2)+√(r²-1)][(r+2)-√(r²-1)] = (r+2)² - (r²-1) = r² + 4r + 4 - r² + 1 = 4r + 5</p><p><strong>Step 2: Simplify Tr</strong></p><p>The (4r+5) terms cancel:</p><p>T_r = √r·√(r+1)·[(r+2)-√(r²-1)]</p><p><strong>Step 3: Decompose into telescoping form</strong></p><p>Notice that: √r·√(r+1)·(r+2) - √r·√(r+1)·√(r²-1) can be split</p><p>After careful manipulation: T_r = [√r·√(r+1)·(r+2)] - [√(r-1)·√r·(r+1)]</p><p>This is telescoping: T_r = √r(r+1)[(r+2) - √(r-1)]</p><p><strong>Step 4: Sum the telescoping series</strong></p><p>∑(r=1 to 16) T_r telescopes to: √16·√17·18 - √0·√1·2</p><p>= 4·√17·18 - 0 = 72√17</p><p><strong>Step 5: Final calculation</strong></p><p>∑T_r/(√68) = 72√17/(√68) = 72√17/(2√17) = 36</p><p><strong>∴ Answer: 36</strong></p>
Correct Answer: 36