Hi
Well we all understood that series is \(\frac{1}{2}\) , \(\frac{-1}{2^2}\) , \(\frac{1}{2^3}\), \(\frac{-1}{2^4}\), \(\frac{1}{2^5}\) ,\(\frac{-1}{2^6}\),\(\frac{1}{2^7}\) ,\(\frac{-1}{2^8}\) ,\(\frac{1}{2^9}\) ,\(\frac{-1}{2^10}\)
Can i rewrite the series as
{\(\frac{1}{2}\) +\(\frac{1}{2^3}\) +\(\frac{1}{2^5}\) +\(\frac{1}{2^7}\) +\(\frac{1}{2^9}\) } - { \(\frac{1}{2^2}\) + \(\frac{1}{2^4}\) + \(\frac{1}{2^6}\)+ \(\frac{1}{2^8}\) + \(\frac{1}{2^10}\) }
{\(\frac{1}{2}\) +\(\frac{1}{2^3}\) +\(\frac{1}{2^5}\) +\(\frac{1}{2^7}\) +\(\frac{1}{2^9}\) } -\(\frac{1}{2}\) { \(\frac{1}{2}\) +\(\frac{1}{2^3}\) +\(\frac{1}{2^5}\) +\(\frac{1}{2^7}\) +\(\frac{1}{2^9}\) }
Let {\(\frac{1}{2}\) +\(\frac{1}{2^3}\) +\(\frac{1}{2^5}\) +\(\frac{1}{2^7}\) +\(\frac{1}{2^9}\) } =q
then
\(q-\frac{q}{2}\)
\(\frac{q}{2}\)
now q is in GP and we need to find the sum of GP whose first term is \(\frac{1}{2}\) and r is \(\frac{1}{4}\) , n is 5
Either you can use the formula for sum of infinte terms of GP or genereal formula .
Lets try the general formula which says sum of n terms of GP
we have \(\frac{\frac{1}{2} * (1 - \frac{1}{4^5})}{ (1-\frac{1}{4})}\)
\(\frac{\frac{1}{2} * ( \frac{4^5-1}{4^5})}{ (\frac{3}{4})}\)
Can we say \(\frac{4^5-1}{4^5}\) is nearly equal to 1
So this expression \(\frac{\frac{1}{2} * ( \frac{4^5-1}{4^5})}{ (\frac{3}{4})}\) hence simplifies to \(\frac{2}{3}\)
now q= \(\frac{2}{3}\)
we need the value of \(\frac{q}{2}\) = \(\frac{1}{3}\)
So Choice D suits our Answer.
PS: Does this solution look long - Yeah because i tried to explain each step. I am sure we will not being many steps that i wrote down . I did get my solution in 1: 03 min
abhimahnaI guess you missed on signs in this post , lucky you ended getting the right answer.
https://gmatclub.com/forum/kth-term-of- ... l#p1724824