Question:

<p><strong>Step 1: Rationalize the general term</strong></p><p>The general term is:<br/>$$a_r = \frac{\sqrt{t^{2r-3}}(1-t)}{(\sqrt{t^{2r-1}}+1)(\sqrt{t^{2r-3}}+1)}$$</p><p>Rationalize by multiplying numerator and denominator by $(\sqrt{t^{2r-1}}-1)(\sqrt{t^{2r-3}}-1)$:</p><p>Denominator becomes: $(t^{2r-1}-1)(t^{2r-3}-1)$</p><p><strong>Step 2: Factor and simplify</strong></p><p>Note that $t^{2r-1}-1 = (t^{2r-3}-1)t^2 + (t^2-1)$</p><p>Actually, use partial fractions. Rewrite:</p><p>$$a_r = \frac{\sqrt{t^{2r-3}}(1-t)}{(\sqrt{t^{2r-1}}+1)(\sqrt{t^{2r-3}}+1)} = \frac{1}{\sqrt{t^{2r-3}}+1} - \frac{1}{\sqrt{t^{2r-1}}+1}$$</p><p><strong>Step 3: Recognize telescoping structure</strong></p><p>The sum becomes:</p><p>$$\sum_{r=2}^{n}\left(\frac{1}{\sqrt{t^{2r-3}}+1} - \frac{1}{\sqrt{t^{2r-1}}+1}\right)$$</p><p>This telescopes: most terms cancel, leaving:</p><p>$$S_n = \frac{1}{\sqrt{t^{1}}+1} - \frac{1}{\sqrt{t^{2n-1}}+1}$$</p><p><strong>Step 4: Take the limit as n→∞</strong></p><p>For $0 < t < 1$: $\lim_{n \to \infty} t^{2n-1} = 0$</p><p>$$P(t) = \frac{1}{\sqrt{t}+1} - \frac{1}{0+1} = \frac{1}{\sqrt{t}+1} - 1 = \frac{1-\sqrt{t}-1}{\sqrt{t}+1} = \frac{-\sqrt{t}}{\sqrt{t}+1}$$</p><p><strong>Step 5: Evaluate at t = 1/2</strong></p><p>$$P\left(\frac{1}{2}\right) = \frac{-\sqrt{1/2}}{\sqrt{1/2}+1} = \frac{-1/\sqrt{2}}{1/\sqrt{2}+1}$$</p><p>$$= \frac{-1/\sqrt{2}}{(1+\sqrt{2})/\sqrt{2}} = \frac{-1}{1+\sqrt{2}}$$</p><p>Rationalize: $$= \frac{-1(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})} = \frac{-(1-\sqrt{2})}{1-2} = \frac{-(1-\sqrt{2})}{-1} = \sqrt{2}-1$$</p><p><strong>Step 6: Check which interval contains $\sqrt{2}-1 \approx 0.414$</strong></p><p>Note: $\sin(\pi/6) = 0.5$, $\tan(\pi/4) = 1$</p><p>Since $0 < 0.414 < 0.5$, we have $P(1/2) \in (0, \sin\pi/6)$</p><p><strong>∴ Answer: A</strong></p>