Prove that: `sum_(m=1)^ntan^(-1)((2m)/(m^4+m^2+2))=tan^(-1)((n^2+n)/(n^2+n+2))`

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

sum_(m=1)^(n) tan^(-1) ((2m)/(m^(4) + m^(2) + 2)) is equal to

Prove that underset(r=1)overset(n)sum "tan"^(-1)(2r)/(r^(4)+r^(2)+2)="tan"^(-1)(n^(2)+n+1)-(pi)/(4)

Prove that sum_(r=0)^(2n) r.(""^(2n)C_(r))^(2)= 2n.""^(4n-1)C_(2n-1) .

Let a_1,a_2 ,a_3 ...... be an A.P. Prove that sum_(n=1)^(2m)(-1)^(n-1)a_n^2=m/(2m-1)(a_1^2-a_(2m)^2) .

The value of ["tan"^(-1)(m)/(n)-tan^(-1)((m-n)/(m+n))] is -

m^(3)-n^(3)-m(m^(2)-n^(2))+n(m-n)^(2)

prove that sum_(r=0)^n(-1)^r^n C_r . [ 1/(2^r)+(3^r)/(2^(2r))+(7^r)/(2^(3r))+(15^r)/(2^(4r))+ ......up to m terms ] = (2^(m n)-1)/(2^(m n)(2^n-1))

IfI_(m , n)=int_0^(pi/2)sin^m xcos^n xdx , Then show that I_(m , n)=(m-1)/(m+n)I_(m-2,n)(m ,n in N) Hence, prove that I_(m , n)=f(x)={((n-1)(n-3)(m-5)(n-1)(n-3)(n-5))/((m+n)(m+n-2)(m+n-4))pi/4 when both m and n are even ((m-1)(m-3)(m-5)(n-1)(n-3)(n-5))/((m+n)(m+n-2)(m+n-4))}

Factorise : 2(m^(2)+n^(2))^(2)-3mn(m^(2)+n^(2))-2m^(2)n^(2)