JEE MAIN - Mathematics (2021 - 17th March Morning Shift - No. 12)
If cot$$-$$1($$\alpha$$) = cot$$-$$1 2 + cot$$-$$1 8 + cot$$-$$1 18 + cot$$-$$1 32 + ...... upto 100 terms, then $$\alpha$$ is :
1.02
1.03
1.01
1.00
Explanation
$${\cot ^{ - 1}}(\alpha ) = co{t^{ - 1}}2 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 + {\cot ^{ - 1}}32 + ....100$$ terms
$$ = {\tan ^{ - 1}}{1 \over 2} + {\tan ^{ - 1}}{1 \over 8} + {\tan ^{ - 1}}{1 \over {18}} + {\tan ^{ - 1}}{1 \over {32}} + ....100$$ term
$$ = \sum\limits_{k = 1}^{100} {{{\tan }^{ - 1}}{1 \over {2{k^2}}}} $$
$$ = \sum\limits_{k = 1}^{100} {{{\tan }^{ - 1}}{2 \over {4{k^2}}} = \sum\limits_{k = 1}^n {{{\tan }^{ - 1}}{{(2k + 1) - (2k - 1)} \over {1 + (2k - 1)(2k + 1)}}} } $$
$$ = \sum\limits_{k = 1}^{100} {\left( {{{\tan }^{ - 1}}(2k + 1) - {{\tan }^{ - 1}}(2k - 1)} \right)} $$
$$ = {\tan ^{ - 1}}201 - {\tan ^{ - 1}}1$$
$$ = {\tan ^{ - 1}}{{200} \over {202}}$$
$$ = {\cot ^{ - 1}}(1.01)$$
Hence $$\alpha = 1.01$$
$$ = {\tan ^{ - 1}}{1 \over 2} + {\tan ^{ - 1}}{1 \over 8} + {\tan ^{ - 1}}{1 \over {18}} + {\tan ^{ - 1}}{1 \over {32}} + ....100$$ term
$$ = \sum\limits_{k = 1}^{100} {{{\tan }^{ - 1}}{1 \over {2{k^2}}}} $$
$$ = \sum\limits_{k = 1}^{100} {{{\tan }^{ - 1}}{2 \over {4{k^2}}} = \sum\limits_{k = 1}^n {{{\tan }^{ - 1}}{{(2k + 1) - (2k - 1)} \over {1 + (2k - 1)(2k + 1)}}} } $$
$$ = \sum\limits_{k = 1}^{100} {\left( {{{\tan }^{ - 1}}(2k + 1) - {{\tan }^{ - 1}}(2k - 1)} \right)} $$
$$ = {\tan ^{ - 1}}201 - {\tan ^{ - 1}}1$$
$$ = {\tan ^{ - 1}}{{200} \over {202}}$$
$$ = {\cot ^{ - 1}}(1.01)$$
Hence $$\alpha = 1.01$$
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