(A) 
(B) 
(C) 
+ 
(D) 
– 2
(E) 
– 2
In this problem many students are tempted to simply add the exponents together and choose answer choice (B).
, however, is not the answer. Were we multiplying the 2’s together then we would add the exponents. The general rule is:
if the bases are the same then add the exponents, only if you are multiplying
In this case, the bases — the number 2 — are the same. So what do we if the bases are being added, not multiplied?
There is no general rule here if you are adding the same bases with different exponents. This problem actually requires that you identify a pattern. When looking for a pattern we want to start with the lower numbers.
+ 
= 6
+ + 
= 14
+ + + 
= 30
You may notice that the sum of each of the series above is very close to the next 2 added:
+ + = – 2
Continuing this pattern we get:
+ + + = – 2
So for the original problem the greatest exponent we are adding is . Therefore, the answer to the original question is – 2. Answer choice (E).
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These kind of operations are quite common in the study of. comp. science.
Suppose you want to find the max. number that an unsigned int 5bit long can hold:
2(2^0 + 2^1 + 2^2 + 2^3 + 2^4). //The reason why I took out a common two, so that all 5 bits are filled with 1’s;
2(2^5 -1)
2^6 – 2.
I got lost on when you started subtracting 2…
Hi Elizabeth,
That’s why these exponents are so exasperating! What we are trying to do here is to identify a pattern that can help us to answer this question without having to actually do all of the math involved here! One of the first things I like to do here is to look at the answer choices and see what they tell us about the question. In this case, the first two answer choices are basically trick answer choices for those who think that we can add the exponents together. If we don’t fall for that trap, we notice that the next three answer choices all include addition and subtraction. Two of them involve subtraction of 2. All of these provide a clue that we are looking for some sort of pattern in the numbers that can be broken down into one of these expressions.
Chris’s approach here was to break this problem down into pieces and look for the pattern. This would be pretty difficult without any guidance, but the last three answer choices provide us with some starting points. If we can identify a pattern that begins to mimic one of those answer choices, we will know that we’ve done it! Once we play around with the numbers a little bit, we can find the pattern and know our answer. I hope this helps to explain our rationale a little bit 🙂
For those who don’t know how to prove this theorem, here is my proof:
First, add 2^1 + 2^2 + 2^3 … + 2^N with 2, which give us 2^1 + 2^2 + 2^3 … + 2^N + 2, we get 2 * 2^1 + 2^2 + 2^3 + … + 2^N, equals 2^2 + 2^2 + 2^3 + … + 2^N.
2^2 + 2^2 + 2^3 + … + 2^N = 2 * 2^2 + 2^3 + … + 2^N
2 * 2^2 + 2^3 + … + 2^N = 2^3 + 2^3 + … + 2^N
2^3 + 2^3 + … + 2^N = 2 * 2^3 + … + 2^N
2 * 2^3 + … + 2^N = 2^4 + … + 2^N
…
Did you see the pattern? Eventually, 2^1 + 2^2 + 2^3 … + 2^N + 2 will lead us to 2^(N+1). So 2^1 + 2^2 + 2^3 … + 2^N = 2^(N+1) – 2
WARNING!
This pattern only works with base 2!
Thanks for sharing your proof (and the important warning!) 🙂
It can also be solved using the formula for Geometric Progression. It is a one-step solution.
Sum = 2 * (2^5 – 1)/(2 – 1) = 62 = 2^6 – 2
Hey Chris,
I solved this question in almost comparable time by taking out 2 common and using parentheses.
Step 1 => 2(1+2+2^2+2^3+2^4)
Step 2 => 2(1+ 2(1+2+2^2+2^3))
Step 3 => 2(1+ 2(1+ 2(1+2+2^2))
Step 4 => 2(1+ 2(1+ 2(1+2+4)))
Step 5 => 2(1+2(1+2(7)))
Step 6 => 2(1+2(15))
Step 7 => 2(1+30)
Step 8 => 2 * 31 = 62
But clearly the pattern way is better, because this problem can be made a lot harder if it were of the type:
Evaluate: 2+2^2+2^3+2^4……2^987 = ?
Steps 5 to 8 can be done mentally, the only time consuming steps in my approach are Steps 1 to 4.
Hi Karan,
Sure, that worked :). But I would stay away from that pattern even if you can do it mentally, just because there are so many steps that you can easily make a careless mistake. And yes, the GRE would definitely choose something much higher in terms of exponents, 2^220, for example :).
Chris, can you give some tips on questions that require establishing a pattern because they can be quite tricky and recognizing a pattern in short span on time is difficult to say the least.
Hi Muhammad,
How about this…send me a link to a few questions that require you to find a pattern and I will help you approach them. One basic tip I can offer is to write out numbers in order to find a pattern. All too often students will stare at a question hoping a pattern will suddenly emerge.
E
how can u get 21+22=6
Hi Yogesh – It looks like we had an issue with the notation. I’m sorry about the confusion.
It should read:
2^1 + 2^2 = 6
The post should be fixed. Let me know if you have any other questions. Thanks!