Limit where n tends to ♾️ tan {Summation of r=1 to n tan^-1 (1/1+r+r^2)} is equal to:

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JEE MAINS 2021, (24th feb,1st shift) Evaluate lim n tends to infinity tan(sigma r=1 to n(tan^-1(1/. JEE MAINS 2021, (24th feb,1st shift) Evaluate lim n tends to infinity tan(sigma r=1 to n(tan^-1(1/.
Let f(x)=lim n→∞ sum r=0 to n(tan(x/2^(r+1))+tan³(x/2^(r+1))/1-tan²(x/2^(r+1))) Then limx→0(e^x-e^f Let f(x)=lim n→∞ sum r=0 to n(tan(x/2^(r+1))+tan³(x/2^(r+1))/1-tan²(x/2^(r+1))) Then limx→0(e^x-e^f
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Let f(x) = lim n tends to infinity sigma r = 0 to r =n ((tan(x/2^(r+1) )+〖tan〗^3 (x/2^(r+1) ))/(1- Let f(x) = lim n tends to infinity sigma r = 0 to r =n ((tan(x/2^(r+1) )+〖tan〗^3 (x/2^(r+1) ))/(1-
Let f(x)= lim n-----0s sigma n,r=0(tanx/2^r+1)+tan^3(x/2^r+1)/1-tan^2(x/2^r+1).Then lim----= Let f(x)= lim n-----0s sigma n,r=0(tanx/2^r+1)+tan^3(x/2^r+1)/1-tan^2(x/2^r+1).Then lim----=
JEE Main 2021 Solution | SUM(t=1 to infinity) tan^-1(6^r/{2^(2r+1)+3^(2r+1)}) #JEEMAIN2021 JEE Main 2021 Solution | SUM(t=1 to infinity) tan^-1(6^r/{2^(2r+1)+3^(2r+1)}) #JEEMAIN2021