Hi.

I like this question. I would solve it as follows:

$$n=1\ \Rightarrow\ \ \ R_{1+1}-R_1=\left(-\frac{1}{2}\right)^1\ \Rightarrow\ \ R_2-R_1=-\frac{1}{2}.$$ $$n=2\ \Rightarrow\ \ \ R_{2+1}-R_2=\left(-\frac{1}{2}\right)^2\ \Rightarrow\ \ R_3-R_2=\frac{1}{4}.$$ This two equations tell us that: $$R_1>R_2\ \ \ \ and\ \ \ \ \ R_3>R_2.$$ Now, we need to find out the relation between R_1 and R_3. If we add the first two equations we will get: $$-R_1+R_3=-\frac{1}{4}\ \Rightarrow\ \ R_3-R_1=-\frac{1}{4}\ \ \ \Leftrightarrow\ \ R_1>R_3.$$ Hence, we conlcude that the correct order is: $$R_1>R_3>R_2.\ \ \ \ OPTION\ A.$$ The correct answer is the option A .

Is that right? <i class="em em-smiley"></i>

=>
If n = 1, then R2 - R1 = -(1/2) < 0, and we have R1 > R2.
If n = 2, then R3 - R2 = 1/4 > 0, and we have R3 > R2.
Furthermore, R3 - R1 = ( R3 - R2 ) + ( R2 - R1 ) = 1/4 + (-1/2) = -1/4 < 0, and we have R3 < R1.
Thus R1 > R3 > R2.

Therefore, A is the answer.
Answer: A