fastest way to do exponent problems? (gmat prep q)

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by pepeprepa » Sat Jul 19, 2008 11:10 am
You can use this formula
"Sum from k=0 to 8 of" 2^k=(1-2^9)/(1-2)=2^9 - 1
And so it is 2^9
because there is 2^0=1 as first term in the formula and we have 2 instead in the question, so we had 1 to 2^9 - 1

But the general formula is this one
"Sum form k=0 to n of" q^k=((1-(q)^(n+1))/(1-q))
It is awful written like that, try to find it on the internet.

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by egybs » Sat Jul 19, 2008 5:46 pm
I don't really understand what pep wrote... Here's an easier way to do this problem instead of memorizing some formula that may or may not apply to another problem you're working on. It may be a fine formula, but my eyes usually glaze over those kinds of things... why? because you don't really get an understanding of how solution works. Because the GMAT can't simply be answered with a series of formulas, it's important to get what's going on here.

2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 =

First of all, you should be able to narrow the choices down to A or B really quickly since addition won't cause exponents to rise that quickly.

The key here is to look for patterns. You'll notice that 2^7 is half of 2^8 and that 2^6 is half of 2^7. You should also notice that 2 is a half of 2^4. Now, looking at the first two terms, (2+2), you'll find that they are equal to the next term, 2^2. Similarly, the sum of 2+2+2^2 = 2^3, because each term is twice the previous term (excluding the extra 2 at the beginning).

This pattern continues so that any given term is equal to the sum of all previous terms. Based on this, we can say that everything before the 2^8, is equal to 2^8. We can therefore simplify this whole equation to:

2^8+2^8 which can be further simplified to:
2^8(1+1) = 2^8 * 2 = 2^9.

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by pepeprepa » Sun Jul 20, 2008 1:22 am
egybs, your way to answer has a lot of class, more beautiful than a mere formula.
However, I have to say formulas are often very simple and fast and if you do not know how the formula works, you can demonstrate it. The thing is that formulas can be really helpful if you do not see the trick as I and karenmeow didn't see it in this question.
When I see 2^2+2^3+2^4 ... or 1^2+2^2+3^2 ... or 1+2+3+4+5 I have my formulas, that's a very reliable way to solve "sum of sets problems"
If someone is interested you can see the formula there
https://en.wikipedia.org/wiki/Geometric_progression

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by gabriel » Sun Jul 20, 2008 5:10 am
egybs wrote:I don't really understand what pep wrote... Here's an easier way to do this problem instead of memorizing some formula that may or may not apply to another problem you're working on. It may be a fine formula, but my eyes usually glaze over those kinds of things... why? because you don't really get an understanding of how solution works. Because the GMAT can't simply be answered with a series of formulas, it's important to get what's going on here.

2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 =

First of all, you should be able to narrow the choices down to A or B really quickly since addition won't cause exponents to rise that quickly.

The key here is to look for patterns. You'll notice that 2^7 is half of 2^8 and that 2^6 is half of 2^7. You should also notice that 2 is a half of 2^4. Now, looking at the first two terms, (2+2), you'll find that they are equal to the next term, 2^2. Similarly, the sum of 2+2+2^2 = 2^3, because each term is twice the previous term (excluding the extra 2 at the beginning).

This pattern continues so that any given term is equal to the sum of all previous terms. Based on this, we can say that everything before the 2^8, is equal to 2^8. We can therefore simplify this whole equation to:

2^8+2^8 which can be further simplified to:
2^8(1+1) = 2^8 * 2 = 2^9.
Nice answer, but always remember that during the exam it might be a little difficult to spot patterns. This is a very basic question on GP and is easily solvable using the formula.

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by shingik » Fri Jul 08, 2011 8:17 pm
Formulas are only applicable if you know when to use them. egybs is right. GMAT is not designed to be solved using a series of formulas. If that were the case, math majors would not even need to study for it. There are basic things that you need to know but if you find yourself trying to memorize formulas you leading yourself astray.

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by Buix0065 » Sat Jul 09, 2011 3:25 am
This is simple factoring... "(2^1 + 2^1) = 2^1(1+1) = 2^1(2^1) = 2^2"


which gives you the idea: 2^n + 2^n = 2^n+1

So you start with 2^1 + 2^1 = 2^2

now you have, 2^2 + 2^2 = 2^3
...

if you continue on with the pattern, eventually you get to 2^8 + 2^8 = 2^9