# Sum of finite series

The sum of series $$\frac{8}{5} + \frac{16}{65} + ....+\frac{128}{2^{18}+1}$$ is

A) $$\frac{540}{1088}$$

B) $$\frac{1088}{545}$$

C) $$\frac{1001}{500}$$

D) $$\frac{1013}{545}$$

I am unable to figure out the general term of series. The answer given is B. How to figure out the general term and solve the question?

What kind of series is this, is it AP, GP, AGP, or combination of these or neither of these?

I want answer till end, because some of answers below only generalized the series but I have no idea to sum that.

• The series has only five terms. You can work this put directly. – Allawonder Feb 24 at 17:40
• @Allawonder, smart man, I would rather leave the question in exam rather than computing it – Usercomingsoon Feb 24 at 17:45
• Not even if you had a calculator, and all you need do is multiply the denominators and notice which of the options is a factor of this product? – Allawonder Feb 24 at 17:47
• Is this meant to be a finite (presumably five-term) sum, or an infinite series, with more terms after the $128/(2^{18}+1)$? – Barry Cipra Feb 24 at 18:51
• @Usercomingsoon, in that case something is off here. The missing terms would seem to be $32/(2^{10}+1)$ and $64/(2^{14}+1)$, but I don't get $1088/545$ from them; I get something with a much messier denominator. – Barry Cipra Feb 24 at 19:25

• Numerator terms get multiplied by $$2$$ every time.
• Denominator $$5 = 2^2+1$$ and $$65 = 2^6+1$$
• Alright, But how would you solve it, I mean I only know ap, gp, agp to solve. But this thing looks like $\frac{8(2^{r-1})}{2^{4r-2}+1}$ – Usercomingsoon Feb 24 at 17:32