Let `C_n` be a circle of radius `(2/3)^n` centered at the origin for `n = 0, 1, 2,....... and A_n` be the area of the region that is inside the circle `C_(2n)` and outside the circle `C_(2n+1)` for `n = 0, 1, 2,......`.The value of the `sum_(n=0)^oo A_n` equals

8. Let S, be a square of unit area. A circle C_1, is inscribed in S_1, a square S_2, is inscribed in C_1 and so on. In general, a circle C_n is inscribed in the square S_n, and then a square S_(n+1 is inscribed in the circle C_n. Let a_n denote the sum of the areas of the circle C_1,C_2,C_3,.....,C_n then lim_(n->oo) a_n must be

The sum of sum_(n=1)^(oo) ""^(n)C_(2) . (3^(n-2))/(n!) equal

If C_(1),C_(2),C_(3),... is a sequence of circles such that C_(n+1) is the director circle of C_(n) . If the radius of C_(1) is 'a', then the area bounded by the circles C_(n) and C_(n+1) , is

If A_n be the area bounded by the curve y=(tanx^n) ands the lines x=0,\ y=0,\ x=pi//4 Prove that for n > 2. , A_n+A_(n+2)=1/(n+1) and deduce 1/(2n+2)< A_(n) <1/(2n-2)

Let C_(0) be a circle of radius 1. For n>=1 let C_(n) be a circle whose area equals the are of a square inscribed in C_(n-1) Then sum_(i=0)^(oo)Area C_(i) equals (A) pi^(2)(B)(pi-2)/(pi^(2))(C)(1)/(pi^(2))(D)(pi^(2))/(pi-2)

The value of sum_(r=0)^(2n)(-1)^(r)*(""^(2n)C_(r))^(2) is equal to :

{a_n} and {b_n} are two sequences given by a_n=(x)^(1//2^n) +(y)^(1//2^n) and b_(n)=(x)^(1//2^n)-(y)^(1//2^n) for all n in N. The value of a_1 a_2 a_3……..a_n is equal to