MathematicsHard179×since 2002Q4282Let 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)}Sk=r=1∑ktan−1(22r+1+32r+16r). Then limk→∞Sk\mathop {\lim }\limits_{k \to \infty } {S_k}k→∞limSk is equal to :Acot−1(32){\cot ^{ - 1}}\left( {{3 \over 2}} \right)cot−1(23)Bπ2{\pi \over 2}2πCtan^−-−1 (3)Dtan−1(32){\tan ^{ - 1}}\left( {{3 \over 2}} \right)tan−1(23)Check answerSkip