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The value of tan^(-1)4/7+tan^(-1)4/(19)+...

The value of `tan^(-1)4/7+tan^(-1)4/(19)+tan^(-1)4/(39)+tan^(-1)4/(67)+ooe q u a l s` `tan^(-1)1+tan^(-1)1/2+tan^(-1)1/3` `tan^(-1)+cot^(-1)3` `cot^(-1)1+cot^(-1)1/2cot^(-1)1/3` `cot^(-1)1+tan^(-1)3`

A

`tan^(-1) 1 + tan^(-1).(1)/(2) + tan^(-1).(1)/(3)`

B

`tan^(-1) 1 + cot^(-1) 3`
`cot^(-1) + cot^(-1).(1)/(2) + cot^(-1).(1)/(3)`

C

`cot^(-1) 1 + cot^(-1).(1)/(2) + cot^(-1).(1)/(3)`

D

`cot^(-1) 1 + tan^(-1) 3`

Text Solution

Verified by Experts

The correct Answer is:
B

Let us find the value of general term of the series
`S = 7 + 19 + 39 + 67 + .. + T_(n)`
`(S = " " 7 + 19 + 39 +.. + T_(n-1) + T_(n))/(0 = 7 + 12 + 20 + 28 + ...(n "terms") -T_(n))` (subtracting)
`:. T_(n) = 7 + ((n -1))/(2) [24 + 8 (n-2)] = 4n^(2) + 3`
`:. T_(n) = tan^(-1).(4)/(4n^(2) + 3) = tan^(-1).(1)/(n^(2) + (3)/(4))`
`= tan^(-1).(1)/(1 + (n^(2) -(1)/(4)))`
`= tan^(-1) [((n+ (1)/(2)) - (n-(1)/(2)))/(1+(n+(1)/(2)) (n-(1)/(2)))]`
`= tan^(-1) (n+(1)/(2)) - tan^(-1) (n -(1)/(2))`
Hence,
`:. S_(n) = underset(r=1)overset(n)sum T_(r) = tan^(-1) (n + (1)/(2)) - tan^(-1).(1)/(2)`
`:. S_(n) = tan^(-1) oo - tan^(-1).(1)/(2)`
`= (pi)/(2) - tan^(-1).(1)/(2)`
`= tan^(-1) 1 + cot^(-1) 3`
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