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by manik11 » Tue Dec 08, 2015 5:02 am
If the infinite sum 1/2^1 + 1/2^2 + 1/2^3 + 1/2^4 +...= 1, what is the value of the infinite sum 1/2^1 + 2/2^2 + 3/2^3 + 4/2^4+....?
A) 1
B) 2
C) 3
D) π (pi)
E) Infinite

OA : B
Source : Veritas Prep

Went completely blank on this one. Experts...could you guys please help me break this question down?
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by GMATGuruNY » Tue Dec 08, 2015 6:10 am
Given:
1/2¹ + 1/2² + 1/2³ + 1/2� + ... = 1

1/2 + 1/4 + 1/8 + 1/16 + ... = 1

15/16 + (sum of increasingly small fractions) = 1

(almost 1) + (sum of increasingly small fractions) = 1.

Implication:
The value in orange is just enough to bring the sum of the lefthand side to 1.
Question:
1/2¹ + 2/2² + 3/2³ + 4/2� + 5/2� + 6/2� + 7/2�...?
= 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + 6/64 + 7/128...

= 1/2 + 1/2 + 3/8 + 2/8 + 5/32 + 3/32 + 7/128...

= 1 + 5/8 + 8/32 + 7/128...

= 1 + 5/8 + 1/4 + 7/128...

= 1 + (almost 1) + (sum of increasingly small fractions)

= 1 + 1

= 2.

The correct answer is B.
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by manik11 » Tue Dec 08, 2015 6:28 am
GMATGuruNY wrote:
manik11 wrote:If the infinite sum 1/2^1 + 1/2^2 + 1/2^3 + 1/2^4 +...= 1, what is the value of the infinite sum 1/2^1 + 2/2^2 + 3/2^3 + 4/2^4+....?
A) 1
B) 2
C) 3
D) π (pi)
E) Infinite
BALLPARK.

1/2¹ + 2/2² + 3/2³ + 4/2� + 5/2� + 6/2� + 7/2�...

= 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + 6/64 + 7/128...

= 1/2 + 1/2 + 3/8 + 2/8 + 5/32 + 3/32 + 7/128...

= 1 + 5/8 + 8/32 + 7/128...

= 1 + 5/8 + 1/4 + 7/128...

= 1 + (almost 1) + (sum of EXTREMELY small fractions)

= 2.

The correct answer is B.
Thanks Mitch!
I have a follow up question regarding this prompt. Why are we given this statement in the question :
"If the infinite sum 1/2^1 + 1/2^2 + 1/2^3 + 1/2^4 +...= 1". I see that you didn't use this in your solution. Am I missing something here?

Thanks!
Manik
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by GMATGuruNY » Tue Dec 08, 2015 6:32 am
manik11 wrote: Thanks Mitch!
I have a follow up question regarding this prompt. Why are we given this statement in the question :
"If the infinite sum 1/2^1 + 1/2^2 + 1/2^3 + 1/2^4 +...= 1". I see that you didn't use this in your solution. Am I missing something here?

Thanks!
Manik
The information in red is not needed to solve the problem.
That said, I've amended my post to show how this information might prove helpful to some test-takers.
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by DavidG@VeritasPrep » Fri Dec 11, 2015 7:45 am
manik11 wrote:
If the infinite sum 1/2^1 + 1/2^2 + 1/2^3 + 1/2^4 +...= 1, what is the value of the infinite sum 1/2^1 + 2/2^2 + 3/2^3 + 4/2^4+....?
A) 1
B) 2
C) 3
D) π (pi)
E) Infinite
One alternative:

We're asked to find the value of: 1/2¹ + 2/2² + 3/2³ + 4/2� ... =1/2 + 2/4 + 3/8 + 4/16..

Call the series we're trying to solve 'x'. So 1/2 + 2/4 + 3/8 + 4/16... = x
If we multiply that series by (1/2) we get 1/4 + 2/8 + 3/16... = (1/2)x

Subtract the second series from the first to get [1/2 + 2/4 + 3/8 + 4/16...] - [1/4 + 2/8 + 3/16...] = 1/2 + 1/4 + 1/8 +1/16 ...
Put another way: x - (1/2)x = 1/2 + 1/4 + 1/8 + 1/16 ...

Note that we're told that 1/2 + 1/4 + 1/8 + 1/16... = 1

So x - (1/2)x =1
(1/2)x = 1
x = 2
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by Matt@VeritasPrep » Fri Dec 11, 2015 2:10 pm
Here's my approach.

We're told that

1/2¹ + 1/2² + 1/2³ + 1/2� + ... = 1

We're asked for

1/2¹ + 2/2² + 3/2³ + 4/2� + ...

which is really

(1/2¹ + 1/2² + 1/2³ + 1/2� + ...) + (1/2² + 1/2³ + 1/2� + ...) + (1/2³ + 1/2� + ...) + (1/2� + ...)

We know that the part in red = 1.

We know that the part in green = the part in red - 1/2
= 1 - 1/2 = 1/2.

We know that the part in blue = the part in green - 1/4
= 1/2 - 1/4 = 1/4.

From here, you can see that we suddenly have 1 + 1/2 + 1/4 + ...,

which is just 1 + (1/2 + 1/4 + ...),

or 1 + (the part in red),

or 1 + 1, or 2.
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