If `S_(n)=1+(1)/(2)+(1)/(2^(2))+"..."+(1)/(2^(n-1))`, then calculate the least value of n such that `2-S_(n)lt(1)/(100)`.

If S_(n) = 1 + (1)/(2) + (1)/(2^(2))+….+(1)/(2^(n-1)) , find the least value of n for which 2- S_(n) lt (1)/(100) .

If S_n = 1+1/2 + 1/2^2+...+1/2^(n-1) and 2-S_n < 1/100, then the least value of n must be :

S_(1),S_(2), S_(3),...,S_(n) are sums of n infinite geometric progressions. The first terms of these progressions are 1, 2^(2)-1, 2^(3)-1, ..., 2^(n) – 1 and the common ratios are (1)/(2),(1)/(2^(2)),(1)/(2^(3)), ...., (1)/(2^(n)) . Calculate the value of S_(1), +S_(2),+ ... + S_(n).

If "^(2n+1)P_(n-1):^(2n-1)P_n=3:5, then find the value of n .

If S_(n)=(1^(2)-1+1)(1!)+(2^(2)-2+1)(2!)+...+(n^(2)-n+1)(n!) , then S_(50)=

If s_n=(1-4/1)(1-4/9)(1-4/25)......(1-4/((2n-1)^2)), where n in N, then

If Delta_(r) = |(1,r,2^(r)),(2,n,n^(2)),(n,(n(n+1))/(2),2^(n+1))| , then the value of sum_(r=1)^(n) Delta_(r) is

If S_(n) = sum_(n=1)^(n) (2n + 1)/(n^(4) + 2n^(3) + n^(2)) then S_(10) is less then

If D_(r) = |(r,1,(n(n +1))/(2)),(2r -1,4,n^(2)),(2^(r -1),5,2^(n) -1)| , then the value of sum_(r=1)^(n) D_(r) , is

If S_(1), S_(2), S_(3),...,S_(n) are the sums of infinite geometric series, whose first terms are 1, 2, 3,.., n and whose common rations are (1)/(2), (1)/(3), (1)/(4),..., (1)/(n+1) respectively, then find the values of S_(1)^(2) + S_(2)^(2) + S_(3)^(2) + ...+ S_(2n-1)^(2) .