Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

m=1n tan-1 [2m] / [m4 + m2 + 2] =

1) tan-1 [2m] / [n2 + n + 2]

2) tan-1 [n2 + n] / [n2 + n + 2]

3) tan-1 [n2 + n + 2] / [n2 + n]

4) None of these

Solution: (2) tan-1 [n2 + n] / [n2 + n + 2]

m=1n tan-1 [2m] / [m4 + m2 + 2]

= ∑m=1n tan-1 [2m] / [1 + (m2 + m + 1) (m2 – m + 1)]

= ∑m=1n tan-1 [(m2 + m + 1) – (m2 – m + 1)] / [1 + (m2 + m + 1) (m2 – m + 1)]

= ∑m=1n tan-1 [(m2 + m + 1) – tan-1 (m2 – m + 1)]

= tan-1 3 – tan-1 1 + tan-1 7 – tan-1 3 + tan-1 13 – tan-1 7 + ….. [tan-1 (n2 + n + 1) – tan-1 (n2 – n + 1)]

= tan-1 (n2 + n + 1) – tan-1 1

= tan-1 [n2 + n] / [2 + n2 + n]

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