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JEE Mains · Maths · STD 11 - 12. limits
Let \(S_{k}=\sum_{r=1}^{k} \tan ^{-1}\left(\frac{6^{r}}{2^{2 r+1}+3^{2 r+1}}\right) .\) Then \(\lim _{k \rightarrow \infty} S_{k}\) is equal to
- A \(\tan ^{-1}\left(\frac{3}{2}\right)\)
- B \(\frac{\pi}{2}\)
- C \(\cot ^{-1}\left(\frac{3}{2}\right)\)
- D \(\tan ^{-1}(3)\)
Answer & Solution
Correct Answer
(C) \(\cot ^{-1}\left(\frac{3}{2}\right)\)
Step-by-step Solution
Detailed explanation
\(S_{ k }=\sum_{ r =1}^{ k } \tan ^{-1}\left(\frac{6^{ r }}{2^{2 r +1}+3^{2 r +1}}\right)\) Divide by \(3^{2 r }\) \(\sum_{r=1}^{k} \tan ^{-1}\left(\frac{\left(\frac{2}{3}\right)^{r}}{\left(\frac{2}{3}\right)^{2 r} \cdot 2+3}\right)\)…
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