If `S_n=cot^(-1)(3)+cot^(-1)(7)+cot^(-1)(13)+cot^(-1)(21)+ddotn` terms, then `S_(10)=tan^(-1)5/6` `S_oo=pi/4` (c) `S_6=sin^(-1)4/5` (d) `S_(20)=cot^(-1)1. 1`

Options

A. `S_(10) = tan^(-1).(5)/(6)`
B. `S_(oo) = (pi)/(4)`
C. `S_(6) = sin^(-1).(4)/(5)`
D. `S_(20) = cot^(-1) 1.1`

Answer

A, B, D

Explanation

Let `t_(r)` denote the `rth` term of the series 3, 7, 13, 21, ... and
`{:(S = 3 + 7 + 13 + 21 + ...+ t_(n)),(S " " 3 + 7 + 13 + ...+ t_(n-1) + t_(n)),(-),(bar(0 = 3 + 4 + 6 + 8 + ..+ 2n - t_(n))):}`
`rArr t_(n) = 3+4 + 6 + .. + 2n = 1 + 2 xx (1)/(2) n (n + 1) = n^(2) + n + 1`
Let `T_(r) = cot^(-1) (r^(2) + r + 1)`
`= tan^(-1) ((1)/(r^(2) + r + 1)) = tan^(-1) ((r + 1 -r)/(1 + r(r + 1))) = tan^(-1) (r + 1) - tan^(-1) r`
Thus, the sum of the first `n` terms of the given series is
`underset(r=1)overset(n) sum [tan^(-1)(r + 1) - tan^(-1) r] = tan^(-1) (n + 1)- tan^(-1) (1)`
`= tan^(-1) [(n + 1 -1)/(1 + 1 (n + 1))]`
`= tan^(-1) ((n)/(n + 2))`
`= tan^(-1) ((1)/(1+ (2)/(n)))`
`rArr S_(oo) = underset(n rarroo)("lim") tan^(-1) ((1)/(1 + (2)/(n))) = (pi)/(4)`,
`S_(10) = tan^(-1).(10)/(12) = tan^(-1). (5)/(6)`
`S_(6) = tan^(-1).(3)/(4) = sin^(-1).(3)/(5)`
`S_(20) = tan^(-1).(10)/(11) = cot^(-1) 1.1`