The question is legit

See here for more https://gre.myprepclub.com/forum/gre-quant ... tml#p54127

Each term after the first is twice the preceding term.

First 10 terms:
1, 2, 4, 8, 16, 32, 64, 128, 256, 512
Since the first term is ODD and remaining terms are all EVEN, the sum of the first 10 terms = ODD + EVEN = ODD

11th and 12th terms:
1024, 2048
Sum = EVEN + EVEN = EVEN

Difference between the second sum and the first sum = EVEN - ODD = ODD
Since the correct answer must be ODD, eliminate B and D.

Rather than calculate, BALLPARK the difference between the two sums:
11th term + 12th term ≈ 1000 + 2000 = 3000
Sum of the first 10 terms ≈ 500 + 250 + 125 + (more than 100) ≈ 1000
Thus:
Approximate difference between the sums = 3000 - 1000 = 2000
Of the remaining answer choices, only E is viable.

Even if you spot the pattern, do you still have to take the time to add all the terms up and then get the difference? Cause that took me a while. Is there a trick or quicker way?

Not really a specific shortcut in this case

The recursive definition \(a_n=2 \times a_{n-1}\) with first term \(a_1=1\) describes a geometric progression where:

First term \(\left(a_1\right)$ : 1\)
Common ratio \((r) : 2\)


The general term for any n -th term in this sequence is given by:


\(a_n=a_1 \times r^{n-1}=1 \times 2^{n-1}=2^{n-1}\)



Calculate the Sum of the First 10 Terms \(\left(S_{10}\right)\)

The sum of the first n terms of a geometric series is calculated using the formula:


\(S_n=\frac{a_1\left(r^n-1\right)}{r-1}\)


For n=10 :


\(S_{10}=\frac{1\left(2^{10}-1\right)}{2-1}=2^{10}-1\)


Since \(2^{10}=1024\) :


\(S_{10}=1024-1=1023\)


The recursive definition \(a_n=2 \times a_{n-1}\) with first term \(a_1=1\) describes a geometric progression where:

First term \(\left(a_1\right) : 1\)
Common ratio \((r) : 2\)


The general term for any n -th term in this sequence is given by:


\(a_n=a_1 \times r^{n-1}=1 \times 2^{n-1}=2^{n-1}\)



Calculate the Sum of the First 10 Terms \(\left(S_{10}\right)\)

The sum of the first $n$ terms of a geometric series is calculated using the formula:


\(S_n=\frac{a_1\left(r^n-1\right)}{r-1}\)


For n=10 :


\(S_{10}=\frac{1\left(2^{10}-1\right)}{2-1}=2^{10}-1\)


Since \(2^{10}=1024\) :


\(S_{10}=1024-1=1023\)