Wild About Math! Making Math fun and accessible

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!

Comments (2) Trackbacks (0)
  1. 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?

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


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