Crazy Comparing

The first thing that should strike (lit. scream at) you is that A is nothing but the difference of the sum of cubes of natural numbers upto (x+1) and x i.e

$\displaystyle A = \sum_{i=0}^{(x+1)} (i+1)^{3} - \sum_{i=0}^x i^3 = (x+1)^3 = x^3 + 3x^2 + 3x +1$

Now, B is not such an easy matter as A. Formally expanding out the non-summation terms and simplifying , we get :

$\displaystyle B = \frac{5x^2}{2} -\frac{x}{2} +1 + \frac{\displaystyle \sum_{i=0}^{(x-2)} i^2 -(x-2)\left(\frac{x-1}{2}\right)}{2}$

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Consider the numerator of the fraction containing the summation :

$\displaystyle \sum_{i=0}^{(x-2)} i^2 -(x-2)\left(\frac{x-1}{2}\right)$

Using the result for the sum of squares of natural numbers, this is equal to :

$\displaystyle \frac{\frac{(x-2)(x-1)(2x-3)}{6} - \frac{(x-2)(x-1)}{2}}{2}$

Clearing denominators and factoring out $(x-1)(x-2)$ , we get the numerator to be :

$\displaystyle \frac{(x-1)(x-2)(2x-6)}{12} = \frac{(x-1)(x-2)(x-3)}{6}$

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Hence, B is :

$\displaystyle B = \frac{5x^2}{2} -\frac{x}{2} +1 +\frac{(x-1)(x-2)(x-3)}{6}$

Simplifying this after multiplying by 2, we get :

$\displaystyle 2B = \frac{x^3}{3} + 3x^2 + \frac{8x}{3}$

We have been given that $x > 0$ . Substitute any positive value in the equations for A and 2B. We hence obtain that $\boxed{A>2B}$

Whew, that was a long solution! Hope that helped :)