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Re: sqrt of 2014 + 2*(1+2+3+..... [#permalink]

take \(2014 = x\) & \(2013 = x - 1\)

The calculation will be simpler.

void wrote:
shortcut for this problem....
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Re: sqrt of 2014 + 2*(1+2+3+..... [#permalink]
Quantity A can be expressed as follow:

The sum of consecutive numbers are (n/2)(Last # + First #)
2014 + 2*[(2013/2)(2014)] = 2014^2 and the sqrt for quantity A is 2014

Quantity A, we let x=2014 and x-1=2013, we get:
(sqrt(x)+1) * ((sqrt(x)+sqrt(x-1)) + (sqrt(x)-sqrt(x-1)) - 2
Going through the math, we get:
(sqrt(x)+1)*(2*sqrt(x)-2) = 2x-2 or 2(x-1)=2*2013

Answer is (B)

This way won't work if this comes up on a quantitative section. Any quicker way? What is the key principles here?
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Re: sqrt of 2014 + 2*(1+2+3+..... [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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