The sum of the first \(2n\) terms of \[1,1,2,\frac{1}{2},4,\frac{1}{4},8,\frac{1}{8},16,\frac{1}{16},..\] is

\(2^n+1-2^{1-n}\),

\(2^n+2^{-n}\),

\(2^{2n}-2^{3-2n}\),

\(\frac{2^n-2^{-n}}{3}\).

Firstly, we notice we can separate the terms of this sequence into

\[[1,2,4,8,16,..]+\left[1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},..\right].\]

We can safely do this because we are asked only to find the sum of the first \(2n\) terms.

If we’d been asked to find the sum to infinity of a sequence like this, than separating the sequence into brackets can sometimes give some strange answers!

For example, look what happens if we try to find the sum to infinity of the sequence \[1 -1 + 1-1+1-1+1....\]

We have

\[(1 -1) + (1-1)+(1-1)+(1-1)+...= 0,\] while \[1 +(-1 + 1)+(-1+1)+(-1+1)...= 1.\]

Which is right?

So the sum of \(2n\) terms of the sequence is the combined sum of the first \(n\) terms of each of these two geometric progressions.

We know that given a geometric sequence with first term \(a\) and common ratio \(r\), the sum of the first \(n\) terms \(S_n = \dfrac{a(r^n-1)}{r-1} = \dfrac{a(1-r^n)}{1-r}\).

So the sum of \(2n\) terms is \[\dfrac{1(2^n-1)}{2-1}+\dfrac{1\left(1-\left(\dfrac{1}{2}\right)^n\right)}{1-\dfrac{1}{2}} = 2^n -1+2-2^{1-n} = 2^n +1-2^{1-n}.\]

So the answer is (a).