13Mar/092

## MMM #27: We have a winner!

Random.org has picked Dooglius as the winner of MMM #27. Congratulations, Dooglius!

The answer to the problem is 1/(w^3-w^2-w+1), or 1/((1-w^2)(1-w))

Tanaeem M Moosa had a very clever demonstration. Check this out:

Let

s=1 + (w+w^2) + (w^2+w^3+w^4) + (w^3 + w^4 + w^5+ w^6) + (w^4 + w^5+ w^6 + w^7 + w^8) + ...

so

w*s= (w) + (w^2+w^3)+ (w^3+w^4+w^5)+ (w^4+w^5+w^6+w^7)+...

so, s-w*s=1+w^2+w^4+w^6+w^8

or, s(1-w) = 1/(1-w^2)

so, s=1/((1-w^2)(1-w))

Blinkdagger will have their next contest on Monday so check it out there!

SomebodyMarch 16th, 2009 - 07:08

Can anybody explain:

s-w*s=1+w^2+w^4+w^6+w^8

goes to:

s(1-w) = 1/(1-w^2)

Is this the sum of an algebraic series that they never felt like explaining in school, but rather felt you should just know?

IBYApril 6th, 2009 - 11:28

Wait, how do you get from the third step to the penultimate step?