Prove that `sum_(r=1)^(n) tan^(-1) ((2^(r -1))/(1 + 2^(2r -1))) = tan^(-1) (2^(n)) - (pi)/(4)`

sum_(r=1)^(n)tan^(-1)((2^(r-1))/(1+2^(2r-1))) is equal to

sum_(r=1)^(n)tan^(-1)((2^(r-1))/(1+2^(2r-1))) is equal to:

sum_(r=1)^9 tan^-1(1/(2r^2)) =

sum_(r=1)^(n) tan^(-1)(2^(r-1)/(1+2^(2r-1))) is equal to a) tan^(-1)(2^n) b) tan^(-1)(2)^n-pi/4 c) tan^(-1)(2^(n+1)) d) tan^(-1)(2^(n+1))-pi/4

sum_(r=1)^(n)tan((x)/(2^(r)))sec((x)/(2^(r-1)));r,n in N

sum_(r=1)^(oo)tan^(-1)((2)/(1+(2r+1)(2r-1)))

Find the sum sum_(r =1)^(oo) tan^(-1) ((2(2r -1))/(4 + r^(2) (r^(2) -2r + 1)))

Find the sum sum_(r =1)^(oo) tan^(-1) ((2(2r -1))/(4 + r^(2) (r^(2) -2r + 1)))

Find the sum sum_(r =1)^(oo) tan^(-1) ((2(2r -1))/(4 + r^(2) (r^(2) -2r + 1)))