Verified Solution • 2025–26 Session moderate

QJEEM21Ma5c9e: JEE Main 2021 Mathematics - Class 12 JEEM21Ma5c9e Solution: Let . Then is equal to :

Let Sk=∑r=1ktan⁡−1(6r22r+1+32r+1){S_k} = \sum\limits_{r = 1}^k {{{\tan }^{ - 1}}\left( {{{{6^r}} \over {{2^{2r + 1}} + {3^{2r + 1}}}}} \right)} . Then lim⁡k→∞Sk\mathop {\lim }\limits_{k \to \infty } {S_k} is equal to :

Multiple Choice Options

  • A cot⁡−1(32){\cot ^{ - 1}}\left( {{3 \over 2}} \right)
  • B π2{\pi \over 2}
  • C tan −- 1 (3)
  • D tan⁡−1(32){\tan ^{ - 1}}\left( {{3 \over 2}} \right)

Step-by-Step Verified Solution

Sk = ∑r=1ktan⁡−1(6r(3−2)(1+(32)2r+1)22r+1)\sum\limits_{r = 1}^k {{{\tan }^{ - 1}}} \left( {{{{6^r}(3 - 2)} \over {\left( {1 + {{\left( {{3 \over 2}} \right)}^{2r + 1}}} \right){2^{2r + 1}}}}} \right)

= ∑r=1ktan⁡−1(2r . 3r+1−3r2r+1(1+(32)2r+1)22r+1)\sum\limits_{r = 1}^k {{{\tan }^{ - 1}}} \left( {{{{2^r}\,.\,{3^{r + 1}} - {3^r}{2^{r + 1}}} \over {\left( {1 + {{\left( {{3 \over 2}} \right)}^{2r + 1}}} \right){2^{2r + 1}}}}} \right)

= ∑r=1ktan⁡−1((32)r+1−(32)r1+(32)r+1(32)r)\sum\limits_{r = 1}^k {{{\tan }^{ - 1}}} \left( {{{{{\left( {{3 \over 2}} \right)}^{r + 1}} - {{\left( {{3 \over 2}} \right)}^r}} \over {1 + {{\left( {{3 \over 2}} \right)}^{r + 1}}{{\left( {{3 \over 2}} \right)}^r}}}} \right)

= ∑r=1k[tan⁡−1(32)r+1−tan⁡−1(32)r]\sum\limits_{r = 1}^k {\left[ {{{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^{r + 1}} - {{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^r}} \right]}

= tan⁡−1(32)2−tan⁡−1(32)1{\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^2} - {\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^1}

+ tan⁡−1(32)3−tan⁡−1(32)2{\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^3} - {\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^2}

+ tan⁡−1(32)4−tan⁡−1(32)3{\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^4} - {\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^3}
.
.
.
+ tan⁡−1(32)k+1−tan⁡−1(32)k{{{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^{k + 1}} - {{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^k}}

= tan⁡−1(32)k+1−tan⁡−1(32)1{{{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^{k + 1}} - {{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^1}}

∴\therefore lim⁡k→∞Sk\mathop {\lim }\limits_{k \to \infty } {S_k}

= lim⁡k→∞[tan⁡−1(32)k+1−tan⁡−1(32)1]\mathop {\lim }\limits_{k \to \infty } \left[ {{{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^{k + 1}} - {{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^1}} \right]

= tan⁡−1(∞)−tan⁡−1(32){{{\tan }^{ - 1}}\left( \infty \right) - {{\tan }^{ - 1}}\left( {{3 \over 2}} \right)}

= π2−tan⁡−1(32){{\pi \over 2} - {{\tan }^{ - 1}}\left( {{3 \over 2}} \right)}

= cot⁡−1(32){\cot ^{ - 1}}\left( {{3 \over 2}} \right)

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