Quote:
Quantity A
\(\sqrt{2014 + 2 * (1+2+3+.....+ 2013)}\)
(1+2+3+.....+ 2013) is a sequence with a common difference 1.
In such arithmetic sequences, the sum = average * number of terms.
\(Average = \frac{1+2013 }{ 2} = \frac{2014 }{ 2}\)
Number of terms = 2013, of course.
\(2 * (1+2+3+.....+ 2013) = 2 * \frac{2014 }{ 2} * 2013 = 2014 * 2013\)
\(2014 + 2 * (1+2+3+.....+ 2013) = 2014 + 2014 * 2013 = 2014 ( 1 + 2013) = 2014^2\)
\(\sqrt{2014 + 2 * (1+2+3+.....+ 2013)} = 2014\)Let's look at another badass expression
Quote:
Quantity B
\(((\sqrt{2014} + 1)^\frac{1}{2} * [(\sqrt{2014} + \sqrt{2013})^\frac{1}{2} - (\sqrt{2014}-\sqrt{2013})^\frac{1}{2}])^2\)
Observe that two parts will squared here.
Apply identity expression (a+b)^2 to the following because it will be wholly squared.
\([(\sqrt{2014} + \sqrt{2013})^\frac{1}{2} - (\sqrt{2014}-\sqrt{2013})^\frac{1}{2}]^2\)
\(= \sqrt{2014} + \sqrt{2013} + \sqrt{2014} - \sqrt{2013} - 2
= 2 \sqrt{2014} - 2\)
The remaining part of the expression is also squared.
\(((\sqrt{2014} + 1)^\frac{1}{2})^2 = \sqrt{2014} + 1\)
Now put this all together.
\(((\sqrt{2014} + 1)^\frac{1}{2} * [(\sqrt{2014} + \sqrt{2013})^\frac{1}{2} - (\sqrt{2014}-\sqrt{2013})^\frac{1}{2}])^2\)
\(= (\sqrt{2014} + 1) * (2\sqrt{2014} - 2)\)
\(= (\sqrt{2014} + 1) * 2* (\sqrt{2014} - 1)\)
Apply identity expression \(a^2 - b^2 = (a+b) (a-b)\),
\(2 * ( \sqrt{2014} + 1) * ( \sqrt{2014} - 1) = 2 * (2014 - 1) = 2 * 2013\)
Therefore, quantity A is 2014, while quantity B is 2*2013.
Hence, B is the correct answer.