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MHTCET 5 May 2025 Shift 2 Paper

Question : 141 of 150

Marks: +1, -0

\cot ^{-1}\left(2 \cdot 1^{2}\right)+\cot ^{-1}\left(2 \cdot 2^{2}\right)+\cot ^{-1}\left(2 \cdot 3^{2}\right)+\ldots \ldots . . \infty=

Solution:

Concept:

Convert each inverse cotangent into inverse tangent and use a telescoping sum.

Explanation:

The given series is

\sum_{r=1}^{\infty} \cot^{-1}(2r^2)

Since

\cot^{-1}x = \tan^{-1}\frac{1}{x}

, we get

\cot^{-1}(2r^2) = \tan^{-1}\left(\frac{1}{2r^2}\right)

Now observe that

\frac{1}{2r^2} = \frac{(1+2r)+(1-2r)}{1-(1+2r)(1-2r)}

Using the tangent addition formula, we have

\tan^{-1}\left(\frac{1}{2r^2}\right) = \tan^{-1}(1+2r) + \tan^{-1}(1-2r)

Substitute

r=1,2,3,\ldots

and add:

\sum_{r=1}^{\infty} \tan^{-1}\left(\frac{1}{2r^2}\right) = \lim_{n \to \infty} \sum_{r=1}^{n} \left[\tan^{-1}(2r+1) - \tan^{-1}(2r-1)\right]

This telescopes to

\lim_{n \to \infty} \left[\tan^{-1}(2n+1) - \tan^{-1}(1)\right]

n \to \infty

\tan^{-1}(\infty) = \frac{\pi}{2}

, and

\tan^{-1}(1) = \frac{\pi}{4}

Therefore, the required sum is

\frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}

Answer:

\frac{\pi}{4}

(Option C).

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