Q29 / 60 · JEE Advanced 2013 (Paper 1) · Mathematics

The value of  cot Σn=123cot-11+Σk=1n2k  is

  • A
    2 3 2 5
  • Correct answer
    B
    2 5 2 3
  • C
    2 3 2 4
  • D
    2 4 2 3

Solution

cot Σn=123cot-11+Σk=1n2k=cotΣn=123cot-11+2+4+6+…+2n
=cotΣcot-11+nn+1
=cotΣtan-1n+1-n1+nn+1 cot-1x=tan-11x
=cotΣn=123tan-1n+1-tan-1ntan-1x-y1+xy=tan-1x-tan-1y
 

Now tan-1n+1-tan-1n=tan-12-tan-11+tan-13-tan-12+....+tan-124-tan-123

=cottan-124-tan-11
=cottan-124-11+24
=cotcot-12523=2523

Hence, the correct answer is option: 'B'.

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