short cut to this??

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by DanaJ » Wed Mar 04, 2009 9:44 pm
There is a formula for smth like this. Consider this a geometrical progression with a = 2 and q = 2. You have the general formula for adding numbers in a geometrical progression:
a + aq + a*q^2 + a*q^3 + ... + a*q^n = a*{[q^(n+1) - 1]/(q - 1)}.

Now, 2 + 2^2 + 2^3 + ... + 2^8 = 2*[(2^9 - 1)/(2 - 1)] = 2*(2^9 - 1) = 2^9 - 2. Then you add the other 2 that is at the beginning to get 2^9 - 2 + 2 = 2^9.

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by coolgirl26 » Wed Mar 04, 2009 11:43 pm
thanks for the response but

you write:


2*[(2^9 - 1)/(2 - 1)] = 2*(2^9 - 1) = 2^9 - 2.

so when 2 *2 ^9 -2
isnt that
2^10, cuz when u multiply u add powers??

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by PAB2706 » Thu Mar 05, 2009 12:24 am
i guess dana has made a small mistake. the rest of her explanation is perfect


the formula is wrong ...It shud be a*{[q^(n) - 1]/(q - 1)}.

this will solve ur problem.

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by kamu » Thu Mar 05, 2009 1:59 am
use a(r^n - 1)/(r-1)

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by DanaJ » Thu Mar 05, 2009 7:28 am
Oops... yes, i did make a mistake, but not where you said I did:
In this case, a = 2, so :
2 + 2^2 + 2^3 + ... + 2^8 = 2(1 + 2 + 2^2 + ... + 2^7) = 2(2^8 - 1) = 2^9 - 2.

The answer should indeed be 2^9, but I just made a silly mistake at the beginning.

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by Brent@GMATPrepNow » Thu Mar 05, 2009 4:41 pm
If you have difficulty memorizing the formula for geometric series and you'd rather not calculate the entire sum from 2 to 2^8, there's one other approach.

Back story:
Finding the sum 2 + 1 + 1/2 + 1/4 + 1/8 + . . . .+ 1/128 + 1/256 + 1/256 is difficult if you don't know the formula for the sum of a finite geometric series.
But, if we reword the question the answer to this related sum becomes clear.
Reworded question: I have 4 kg of gold. On the first day I give you half of my gold (2kg). On the second day, I give you half of what remains (1kg). On the third day, I give you half of what remains (1/2 kg) and so on.
As you can see, the amount of gold in your possession becomes the sum 2 + 1 + 1/2 + 1/4 + 1/8 + . . . .etc
Since I began with 4 kg of gold, this sum slowly gets closer and closer to 4 (kg)
At one point, I have 1/128 kg remaining. The next day, I give you half (1/256) and I give you the other half as well (1/256).
I have now given you all of my gold. (4kg)

Now apply this to the original question.
I have 2^9 kg of gold.
On day one, I give you half of that (2^8 kg).
On day two, I give you half of the remaining 2^8 kg (2^7 kg)
And so on.
You see that we get the same sum (2^8 + 2^7 + 2^6 + . . . +2+2)
Since we began with 2^9 kg of gold, the sum must be 2^9.
Brent Hanneson - Creator of GMATPrepNow.com
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