1.

If `S_n=cot^-1(3)+cot^-1(7)+cot^-1(13)+cot^-1(21)+.....`, `n` terms, then

Answer» `S_n = cot^-1(3)+ cot^-1(7) + cot^-1(13) + cot^-1(21) .....` n terms
`T_n = cot^-1(n^2 +n +1)`
`S_n = sum_(n=1)^n cot^-1(n^2+n+1)`
`= sum_(n=1)^n tan^-1[ 1/(n^2+n+1)]`
`= sum_(n=1)^n tan^-1((n+1-n)/(1+n(n+1)))`
`sum_(n=1)^n tan^-1 (n+1) - tan^-1(n)`
`= tan^-1 2 - tan^-1 1 + tan^-13 ..... tan^-12+ ....+ tan^-1 (n+1) `
`= tan^-1(n+1) - tan^-1(1)`
`S_10 = tan^-1(11) - tan^-1(1)`
`= tan^-1((11-1)/(1+11)) = tan^-1(10/12) = tan^-1(5/6)`
`S_oo = n-> oo { tan^-1 (n+1) - tan^1(1)}`
`pi2 - pi/4 = pi/4`
`S_6 = tan^-1 7 - tan^-1 1= tan^-1((7-1)/(1+7)) = tan^-1(6/8) = tan^-1(3/4)`
`S_20 = tan^-1 21 - tan^-1= tan^-1((21-1)/(1+21)) = tan^-1(20/22)`
option `a, b,d` are correct
answer


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