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# Exasperating Exponents

(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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### 11 Responses to Exasperating Exponents

1. Tian Tuo You July 30, 2016 at 1:06 pm #

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!

• Magoosh Test Prep Expert August 2, 2016 at 6:35 am #

Thanks for sharing your proof (and the important warning!) 🙂

2. Obaid October 24, 2015 at 4:06 pm #

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

3. Karan August 4, 2014 at 1:31 pm #

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 = ?

• Karan August 4, 2014 at 1:38 pm #

Steps 5 to 8 can be done mentally, the only time consuming steps in my approach are Steps 1 to 4.

• Chris Lele August 12, 2014 at 11:19 am #

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 :).

4. Muhammad June 19, 2012 at 9:47 pm #

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.

• Chris June 20, 2012 at 3:18 pm #

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.

5. Mohamed July 12, 2011 at 8:48 am #

E

6. Yogesh Shraddha April 10, 2011 at 9:04 am #

how can u get 21+22=6

• hansoo April 10, 2011 at 5:46 pm #

Hi Yogesh – It looks like we had an issue with the notation. I’m sorry about the confusion.