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amitdgr
\(2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\) = ?

* 2^9
* 2^10
* 2^16
* 2^35
* 2^37

See for pattern:
2 + 2 =2^2
2^2+2^2=2^3
2^3 + 2^3=2^4
........
2^8 + 2^8=2^9

wouldnot take more than 30 seconds.

hope this helps..!!
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cangetgmat
I was stuck performing addition of GP.
+1 to you

You can do it using sum of GP as well.

\(2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = 1 + (1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8)\)
The terms in brackets make a GP with first term = 1 and common ratio = 2

GP = \(1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
Sum of 9 terms of GP \(= 1(1 - 2^9)/(1-2) = 2^9 - 1\)

Sum of the required series = 1 + sum of GP = \(1 + 2^9 - 1 = 2^9\)
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Baten80
\(2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?\)


A. \(2^9\)

B. \(2^{10}\)

C. \(2^{16}\)

D. \(2^{35}\)

E. \(2^{37}\)


don't neglect what might be the easiest way to solve this problem -- and what's absolutely the most straightforward: just convert these to concrete numbers.

the problem statement is
2 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256

the first few terms here are pretty insignificant, as far as the total sum is concerned. 32 + 64 is pretty close to 100, so, this sum will be close to 100 + 128 + 256.

the answer choices are
A/ 512
B/ 1024
C/ ha, you must be kidding
D/ ha, you must be kidding
E/ ha, you must be kidding.

this method certainly doesn't take any MORE time than any of the other methods here... and it may in fact take less, for many people.

Posted from my mobile device
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amitdgr
\(2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\) = ?

* 2^9
* 2^10
* 2^16
* 2^35
* 2^37

See for pattern:
2 + 2 =2^2
2^2+2^2=2^3
2^3 + 2^3=2^4
........
2^8 + 2^8=2^9

wouldnot take more than 30 seconds.

hope this helps..!!

Great explanation kraizada84, one question why is 2^2+2^2=2^3 and not 2^4 ? Same with 2^3 + 2^3=2^4 and not 2^6?....
Thanks for your time in advanced.
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Kimberly77


one question why is 2^2+2^2=2^3 and not 2^4 ? Same with 2^3 + 2^3=2^4 and not 2^6?....
Thanks for your time in advanced.

Recommend studying up on exponents.

2^2 + 2^2 = 2^2(1+1) = 2^2(2)= 2^3

Exponents are added only when two like terms are multiplied

2^2 * 2^2 = 2^4

Posted from my mobile device
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Great thanks Regor60. My silly mistake and make sense now.
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Solving using two methods

Method 1: Logic

\(2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= \(2*2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= \(2^2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= \(2*2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= \(2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
Following the same pattern till last term we will get
.
.
.
= \(2^8 + 2^8\)
= \(2*2^8\)
= \(2^9\)

So, Answer will be A

Method 2: Geometric series formula

\(2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= 2 + a Geometric series whose first term is 2, common ratio is 2 and number of terms is 8
= 2 + 2*\(\frac{2^8 - 1 }{ 2 - 1}\)
= 2 + \(2^9\) - 2
= \(2^9\)

So, Answer will be A
Hope it helps!

Watch the following video to learn How to Sequence problems

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All the methods are great but counting them in my head was the fastest for me lol
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