[" If "S_(n)=1+(1)/(2)+(1)/(2^(2))+...+(1)/(2^(n)-1)" ,then colculate "],[" the Least value of ni such that "2-S_(n)<(1)/(100)" is "]

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 :

If S_(n)=1 + (1)/(2) + (1)/(2) + …..+ (1)/(2^(n-1)), (n in N) then …….

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)) , 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) . Calculate the least value of n such that S_n=2-S_n lt 1/100 .

Let s_(n)=1+(1)/(3)+(1)/(3^(2))+…+(1)/(3^(n-1)) . The least value of n in N such that (3)/(2) -S_(n) lt (1)/(400) is

If S=sum_(n=1)^(10)(2n+(1)/(2)), then S

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 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 :

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 :