limn→∞ ∑k=1n tan−1⁡12k2 is

limnk=1ntan112k2 is

  1. A

    π/8

  2. B

    π/2

  3. C

    π/4

  4. D

    π/3

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    Solution:

    tan112k2=tan121+(2k1)(2k+1)=tan1(2k+1)(2k1)1+(2k1)(2k+1)=tan1(2k+1)tan1(2k1)

    k=1ntan112k2=tan1(2n+1)tan1(2n1)+tan1(2n1)tan1(2n3)++tan1 3tan11=tan1(2n+1)tan11limnk=1ntan112k2=tan1π4=π2π4=π4

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