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InterviewSolution
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The sum of the infinite series `cot^(-1)(7/4)+cot^(-1)((19)/4) +cot^(-1)((39)/4)....oo` |
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Answer» Here, general term `T_r` can be wriiten as, `T_r = cot^-1((4r^2+3)/(4)) = cot^-1(r^2+3/4)` `=>T_r = tan^-1(1/(r^2+3/4)) = tan^-1(((r+1/2)-(r-1/2))/(1+r^2-1/4))` `= tan^-1(((r+1/2)-(r-1/2))/(1+(r+1/2)(r-1/2)))` We know, `tan^-1((x-y)/(1+xy)) = tan^-1x+tan^-1y` `:. T_r = tan^-1(r+1/2) - tan^-1(r-1/2)``:. T_1 = tan^-1 (3/2)-tan^-1 (1/2)` `T_2 = tan^-1 (5/2)-tan^-1 (3/2)` `T_3 = tan^-1 (7/2)-tan^-1 (5/2)` `T_n = tan^-1 ((2n+1)/2)-tan^-1 ((2n-1)/2)` So, the required sum will be, `sum T_r = tan^-1((2n+1)/2) - tan^-1(1/2)` `=pi/2-tan^-1(1/2)` `:. sumT_r= cot^-1(1/2)` |
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