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

Answer - A

This problem is killing me. Any thoughts on how to handle this problem?

There is some type of pattern with the number 2 and increasing exponents that I can't quite understand.

shom

There are a number of ways to handle this. One easy solution is to have your powers of two memorized up to 2^10. From there, we can see that we just need to add together: 2 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256. Start at the beginning and add upward. You will notice that as you go you are continually adding a new amount to the amount that you already had. (Do it out loud if you need to, to see this.) From that, we can see that we are really just taking the original 2, and multiplying it by 2 eight more times, for 2^9.

can this be done this way ? I find its time consuming to expand the powers, add and then covert it to power agian.

2+2+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 [ 1+1+2+2^2+2^3+2^4+2^5+2^6]
=2^2 . 2^2 [1+1+2+2^2+2^3+2^4]
= 2^2 . 2^2 .2^2 [1+1+2+2^2 ]
=2^2 . 2^2 .2^2 .2^2 [2]

= 2^9

This can be solved using Sum of GP formulae:

Sum of GP=a(1-r^n)/1-r *When r is less than one.
a(1-r^n)/r-1 *when r is greater than 1.

2+2(2^8-1)/(2-1)=2+2^9-2=2^9.Hence the answer A.

Here is one way:

2 + 2 + 2^2 + 2^3 + ...

Start with grouping the first 2 terms.
2 + 2 = 4 = 2^2.
Now the series becomes
2^2 + 2^2 + 2^3 + ...
Now take 2^2 + 2^2 which is actually 2(2^2) which is 2^3 which you can now combine with the next term.
Continue and you will find that it is 2.2^8 which is 2^9.

-z

gauravbawa's method, directly above, is nice.

there are also nice BACKUP METHODS here.

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first, the humble method of PATTERN RECOGNITION:
FACT:
the gmat will never make you execute the same operation more than three or four times.
if you see a pattern calling for you to execute the same operation an unreasonable number of times, then LOOK FOR A PATTERN.

let's look for a pattern here as we sum the numbers:
first number = 2
sum of first two #s = 4
sum of first three #s = 8
sum of first four #s = 16
ok, i think i get what's going on. let's see: the first # (which is the "sum of the first 1 number") is 2^1; the sum of the first 2 numbers is 2^2; the sum of the first 3 numbers is 2^3; and so on.
there are nine numbers in the list, so you'll get 2^9.
ans = (a)

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then, the humble method of ESTIMATION:
you should know your powers of 2, at least up to 2^8 = 256 (which, conveniently, happens to be the biggest number in the sum).
going on that knowledge, this sum is mostly a sum of VERY small numbers; the only decently big numbers in the list are the last few, namely, 32, 64, 128, and 256.
there is NO WAY that a sum of such numbers could be anything other than (a) or maybe (b), and adding together just those last 4 terms gives 480. therefore, the sum is 480 + small stuff, which looks a lot more like (a) than (b).

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finally, you could easily just FIND THE NUMBERS AND ADD THEM TOGETHER. you should have no trouble doing so within the time constraint of the problem, provided that you START DOING SO IMMEDIATELY.
this is the crux of any "backup" solution method: don't stare at the problem. if you think of something that could work, get started on it right away.