" Let "S=sum_(n=1)^(99)(5^(100))/((25)^(n)+5^(100))*" Find "[S]

Find the value of S_(n)=sum_(n=1)^(oo)(3^(n)*5^(n))/((5^(n)-3^(n))(5^(n-1)-3^(n+1))) and hence s_(oo)

Let t_(100)=sum_(r=0)^(100)(1)/(("^(100)C_(r ))^(5)) and S_(100)=sum_(r=0)^(100)(r )/(("^(100)C_(r ))^(5)) , then the value of (100t_(100))/(S_(100)) is (a) 1 (b) 2 (c) 3 (d) 4

Let t_(100)=sum_(r=0)^(100)(1)/(("^(100)C_(r ))^(5)) and S_(100)=sum_(r=0)^(100)(r )/(("^(100)C_(r ))^(5)) , then the value of (100t_(100))/(S_(100)) is (a) 1 (b) 2 (c) 3 (d) 4

Let S=sum_(n=1)^(9999)(1)/((sqrt(n+1))(root(4)(n)+root(4)(n+1))), then S equals

Let S_(n)=sum_(k=1)^(4n)(-1)^((k(k+1))/(2))*k^(2) , then S_(n) can take value

Let a^(n) be the nth term of the G.P. of positive numbers. Let sum_(n=1)^(100) a_(2n) = alpha and sum_(n=1)^(100) a_(2n-1) = beta , such that alpha cancel(=)beta then the common ration is :

Let a_(n) be the nth term of the G.P of positive numbers . Let sum_(n=1)^(100) a_(2n) = alpha and sum_(n=1)^(100) a_(2n-1) =beta such that alpha ne beta then the common ratio is

S_(n)=sum_(n=1)^(n)(n)/(1+n^(2)+n^(4)) then S_(10)S_(20)