Bunuel wrote:

For every integer n from 1 to 200, inclusive, the nth term of a certain sequence is given by \((-1)^n*2^{(-n)}\). If N is the sum of the first 200 terms in the sequence, then N is

(A) less than -1

(B) between -1 and -1/2

(C) between -1/2 and 0

(D) between 0 and 1/2

(E) greater than 1/2

the sequence can be written as \(\frac{-1}{2}, \frac{1}{2^2}, \frac{-1}{2^3}, \frac{1}{2^4}......................., \frac{-1}{2^{199}}, \frac{1}{2^{200}}\). For simplicity let \(\frac{1}{2}=a\)

so \(N= (a^2+a^4+.....a^{200})-(a+a^3+......a^{199})\). This is a GP series. Sum of GP series is \(a_1*\frac{(1-r^n)}{1-r}\). Here we have two GP series and each has 100 terms, for the first series \(a_1=a^2\) and for the second series \(a_1=a\), common ratio, \(r=a^2\) and \(n=100\), which is same for both the series.

Hence \(N= a^2*\frac{[1-(a^2)^{100}]}{1-a^2} - a*\frac{[1-(a^2)^{100}]}{1-a^2} = \frac{[1-(a^2)^{100}]}{1-a^2}*(a^2-a)\). Now substitute the value of \(a\) here to get

\([(1-\frac{1}{2^{200}})/(1-\frac{1}{2^2})]*(\frac{1}{2^2}-\frac{1}{2})\). on solving this equation, we will get \((\frac{1}{2^{200}}-1)*\frac{1}{3}\)

Among the options given, this value will fall between \(\frac{-1}{2}\) and \(0\)

Option

C