Properties of Log-Sums

We have,

$\displaystyle A = \sum_{n=1}^{2014} n = 1 + 2 + 3 + ... + 2013 + 2014 = \left(\frac{2014(2014+1)}{2}\right) =$

something big $\displaystyle= X$

And,

$\displaystyle B = \sum_{n=1}^{2014} n^3 = 1^3 + 2^3 + 3^3 + .... + 2013^3 + 2014^3 = \left(\frac{2014(2014+1)}{2}\right)^{2} =$ something even bigger $\displaystyle = X^2$

What we need is

$\displaystyle log_{ \sqrt{X}}{X^2} = \frac{2}{1/2}log_{X}{X} = \boxed{4}$

And hence the answer!

Cool fact: the square of a triangular number up to a number n is equal to the sum of the first n cubes!

A= n(n+1)/2 and B= (n(n+1)/2)^2 ,, i.e B=A^2 ,,, (logB)/(log(A)^0.5))=4

A=n(n+1)/2 B={n(n+1)/2}^2 Log√a b =log{ {n(n+1)/2}^0.5 {n(n+1)/2}^2 =4log n(n+1)/2 n(n+1)/2 =4