Question

If $\sum_{r=1}^{50} \tan^{-1}\frac{1}{2r^{2 = p}}$ then $\tan p$ is

• A. $\frac{100}{101}$
• B. $\frac{51}{50}$
• C. $\frac{50}{51}$
• D. $\frac{101}{102}$

Hint:

Recognize patterns that allow for telescoping sums, especially with inverse trigonometric functions. The key is to manipulate the argument of the $\tan^{-1}$ function into the form $\frac{A-B}{1+AB}$ so it can be expressed as $\tan^{-1}A - \tan^{-1}B$.

Answer 1

The Correct Option is A

Step 1: Given \[ p = \sum_{r=1}^{50} \tan^{-1}\left(\frac{1}{2r^2}\right) \] Rewrite: \[ \frac{1}{2r^2} = \frac{2}{4r^2} \] \[ p = \sum_{r=1}^{50} \tan^{-1}\left(\frac{2}{4r^2}\right) \]
Step 2: Use identity \[ \tan^{-1}A - \tan^{-1}B = \tan^{-1}\left(\frac{A-B}{1+AB}\right) \] Let: \[ A = 2r+1,\quad B = 2r-1 \] Then: \[ A-B = 2,\quad 1+AB = 1 + (2r+1)(2r-1) = 4r^2 \] \[ \Rightarrow \frac{A-B}{1+AB} = \frac{2}{4r^2} \]
Step 3: Substitute: \[ p = \sum_{r=1}^{50} \left[\tan^{-1}(2r+1) - \tan^{-1}(2r-1)\right] \]
Step 4: Telescoping: \[ p = (\tan^{-1}3 - \tan^{-1}1) + (\tan^{-1}5 - \tan^{-1}3) + \cdots + (\tan^{-1}101 - \tan^{-1}99) \] Cancelling terms: \[ p = \tan^{-1}(101) - \tan^{-1}(1) \]
Step 5: Use identity again: \[ p = \tan^{-1}\left(\frac{101-1}{1 + 101\cdot1}\right) = \tan^{-1}\left(\frac{100}{102}\right) = \tan^{-1}\left(\frac{50}{51}\right) \]
Step 6: Find $\tan p$: \[ \tan p = \frac{50}{51} \]
Final Answer:
\[ \boxed{\frac{50}{51}} \]