`l_(n)=int_(0)^(pi//4)tan^(n)xdx`, then `lim_(nto oo)n[l_(n)+l_(n-2)]` equals

If l_(n)=int_(0)^(pi//4) tan^(n)x dx, n in N "then" I_(n+2)+I_(n) equals

Let a_(n)=int_(0)^(pi//2)(1-sint )^(n) sin 2t, then lim_(n to oo)sum_(n=1)^(n)(a_(n))/(n) is equal to

If I_(n)=overset(pi//4)underset(0)int tan ^(n) x dx, lim_(n to oo) n(I_(n+1)+I_(n-1)) equals

I_n = int_0^(pi/4) tan^n x dx , then the value of n(l_(n-1) + I_(n+1)) is

I_(n)=int_(0)^(pi//4) tan^(n)x dx , where n in N Statement-1: int_(0)^(pi//4) tan^(4)x dx=(3pi-8)/(12) Statement-2: I_(n)+I_(n-2)=(1)/(n-1)

If l_(n)=int_(0)^((pi)/(4)) tan^(n) xdx show that (1)/(l_(2)+l_(4)),(1)/(l_(3)+l_(5)),(1)/(l_(4)+l_(6)),(1)/(l_(5)+l_(7)),"...." from an AP. Find its common difference.

If the integral l_(n)=int_(0)^(pi//4)tan^(n)xdx is reduced to its lower integrals like l_(n-1),l_(n-2) etc., The value of (l_(3)+2l_(5))/(l_(1)) is

lim_(nto oo) (2^n+5^n)^(1//n) is equal to

lim_(nto oo)sum_(r=1)^(n)r/(n^(2)+n+4) equals

(lim)_(n to oo)n/(2^n)int_0^2x^n dx equals___