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

[agalstia]

Since all the bases are the same, I merely added up all the exponents to equal 37, but that is the wrong answer. I also tried to factor out a 2^2 from each number, but I was lost as to which direction to go afterwards.

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

OA is A. This is from the GMAT Prep software.

[rajivbhatia2007]

I guess there are different ways you could do this problem:
For GMAT, you probably should memorize the powers of 2 up to 2^10. Adding it out

2+2+4+8+16+32+64+128+256

The answer choices:

a) 2^9 = 512
b) 2^10 = 1024
c) 2^16 = (2^6)(2^10) = 64 x 1024 approx 64,000 (by looking at the addition, clearly this number is too big)
d) clearly too big
e) clearly too big

By adding up the choices, the answer is A


The other way to do this is:
2, 2, 4, 8, 16, 32, 64, 128, 256
Each term in the sequence is equal to the sum of the preceding terms. So the term 256 is the sum of 2 + 2 + 4 + .... + 128. Therefore: 2 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 = 256 + 256 = 512 = 2^9

best,
rajiv
my gmat experience: http://gmathints.com/

[agalstia]

thanks. that makes sense. the first approach just seems like it would take longer than 2 minutes.

[jssaggu.tico]

The sum of the first n terms of a geometric progression is:
a(|r^n-1|)/|1 - r| ..................1
for the following type of series

Sn = a + ar + ar2 + ... + ar^(n-1)=a(1+r+r^2+r^3 ... +r^(n-1))

We have 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^7+2^8
=2+2(1+2+2^2+2^3+2^4+.....2^7)
=2+2(2^8-1)=2^9---by using 1

Just remember the series formula, and rest is cup of cake. It really helps when we have much complex geometric series or in case, we have general series with terms of 'n'

[tim]

thanks, Navjot..

[sarahmailings]

The sum of the first n terms of a geometric progression is:
a(|r^n-1|)/|1 - r| ..................1
for the following type of series

Sn = a + ar + ar2 + ... + ar^(n-1)=a(1+r+r^2+r^3 ... +r^(n-1))

We have 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^7+2^8
=2+2(1+2+2^2+2^3+2^4+.....2^7)
=2+2(2^8-1)=2^9---by using 1

Just remember the series formula, and rest is cup of cake. It really helps when we have much complex geometric series or in case, we have general series with terms of 'n'

I'm not quite sure I understand the series formula explanation. I did find some of the explanations posted on this forum quite helpful. Maybe others will too.

[graphica]

It is a bit lengthy but can done.

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

Rewrite the first 2+2 as 2^2

Therefore

2^2[1+1+2+2^2+2^3+2^4+2^5+2^6] Now making 1+1+2=2^2
2^2[2^2{1+1+2+2^2+2^3+2^4}] Now making 1+1+2=2^2
2^2[2^2{2^2(1+1+2+2^2)}] Now making 1+1+2=2^2
2^2[2^2{2^2(2^2+2^2)}] Now making 1+1+2=2^2
2^2[2^2{2^2(2^2(1+1))}] Now making 1+1=2
Finally it comes to
2^2 X 2^2 X 2^2 X 2^2 X 2

Adding the powers since the base is the same makes it
2^2+2+2+2+1 = 2^9

I hope it helps.

[mschwrtz]

Alternatively you could look for a pattern.

2+2=2^2

2+2+2^2=2^3

so you can see that each term in the expression is equal to the sum of all the previous terms, so that adding each term is doubling the previous sum. Doubling=multiplying by 2=increasing the power by 1.

[SandeshM859]

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)+2^3+2^4+2^5+2^6+2^7+2^8
=2^3+2^3+2^4+2^5+2^6+2^7+2^8
=2^3(1+1)+2^4+2^5+2^6+2^7+2^8
=2^4+2^4+2^5+2^6+2^7+2^8
I hope you see a pattern here!
All of this will boil down to 2^8+2^8= 2^9

[RonPurewal]

you can do that.

[RonPurewal]

by the way, 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 (myself included).