Quiz#3 NOT(Rubik's Cube)

[drsarao]

Taurean,
Thanks for opening my eyes!
Fazza's solution is really elegant.
I 'thought' it was a 'series' solution with n expressions. And never even checked it out.
Now, I tried (honest) but could not figure it out. (Algebra is not a very strong point for me!)

I would be delighted if Fazza can throw some light. Thanks.

Array solution: would multiply each integer upto (and including) the given number X to itself. Sum all and then multiply with 6.

Fazza's solution: X*(X+1)*(2X+1) Can't wrap my brain around it.

What would be the solution if there was no multiplication with 6 at the end?

 

[taurean]

Hi drsarao,

Here's one derivation. You will see that the actual series is for calculating SUM of SQUARES on one side of the cube.

The sum of the squares of the first n natural numbers is in its various forms:

So when we had to calculate it for all 6 sides the divisor 6 disappeared.

 

[BiocideJ]

That is absolutely brilliant. I am with drsarao that I believed this could only be accomplished with some form of series expression and didn't realize that fazza's initial response wasn't giving that.

Thank you for the enlightenment.

 

[Fazza]

Shrivallabha, thank you. Best regards, Fazza

 

[drsarao]

Hi drsarao,

Here's one derivation. You will see that the actual series is for calculating SUM of SQUARES on one side of the cube.

The sum of the squares of the first n natural numbers is in its various forms:

So when we had to calculate it for all 6 sides the divisor 6 disappeared.

Plain Nirvana!

Thanks Srivallabha and Fazza.

So the original equation would be: ( X*(X+1)*(2X+1))/6
(multiplying by 6 in fact simplified the equation!)

PS
I honestly tried to follow the explanations in Ken Ward's Mathematics Pages. I have bookmarked it for future revisit!