Question

If \(\sum_{r=1}^{50} \tan^{-1}\frac{1}{2r^2} = p\), then the value of \(\tan p\) is :

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

Hint:

For series involving \(\tan^{-1}\), the main goal is to transform the argument into the form \(\frac{A-B}{1+AB}\). A common trick is to multiply the numerator and denominator by a constant to help create the desired form, as we did by writing \(\frac{1}{2r^2}\) as \(\frac{2}{4r^2}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question
We need to evaluate a sum of inverse tangent functions. This type of problem usually involves creating a telescoping series using the formula \(\tan^{-1}x - \tan^{-1}y = \tan^{-1}\frac{x-y}{1+xy}\).
Step 2: Key Formula or Approach
We will try to express the general term \(\tan^{-1}\frac{1}{2r^2}\) in the form \(\tan^{-1}(A) - \tan^{-1}(B)\).
Let's manipulate the argument \(\frac{1}{2r^2}\). To use the formula, we need a '1' in the denominator. \[ \frac{1}{2r^2} = \frac{2}{4r^2} = \frac{2}{1 + 4r^2 - 1} = \frac{2}{1 + (2r-1)(2r+1)} \] Now, we can write the numerator as a difference of the terms in the product: \((2r+1) - (2r-1) = 2\). So the argument is \(\frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}\).
Step 3: Detailed Explanation
The general term of the series, \(T_r\), is: \[ T_r = \tan^{-1}\frac{1}{2r^2} = \tan^{-1}\left(\frac{(2r+1)-(2r-1)}{1+(2r-1)(2r+1)}\right) \] Using the identity \(\tan^{-1}A - \tan^{-1}B = \tan^{-1}\frac{A-B}{1+AB}\), with \(A=2r+1\) and \(B=2r-1\), we get: \[ T_r = \tan^{-1}(2r+1) - \tan^{-1}(2r-1) \] Now we can write out the sum, which is a telescoping series: \[ p = \sum_{r=1}^{50} T_r = \sum_{r=1}^{50} [\tan^{-1}(2r+1) - \tan^{-1}(2r-1)] \] \[ p = [\tan^{-1}(3) - \tan^{-1}(1)] + [\tan^{-1}(5) - \tan^{-1}(3)] + [\tan^{-1}(7) - \tan^{-1}(5)] + \dots + [\tan^{-1}(101) - \tan^{-1}(99)] \] All the intermediate terms cancel out. We are left with: \[ p = \tan^{-1}(101) - \tan^{-1}(1) \] Now we use the formula again to combine these two terms: \[ p = \tan^{-1}\left(\frac{101-1}{1+101 \times 1}\right) = \tan^{-1}\left(\frac{100}{102}\right) = \tan^{-1}\left(\frac{50}{51}\right) \] Step 4: Final Answer
We are asked to find the value of \(\tan p\). \[ \tan p = \tan\left(\tan^{-1}\left(\frac{50}{51}\right)\right) = \frac{50}{51} \]