MMM #27: We have a winner!

March 13th, 2009 | by Sol |

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!

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  1. 2 Responses to “MMM #27: We have a winner!”

  2. By Somebody on Mar 16, 2009 | Reply

    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?

  3. By IBY on Apr 6, 2009 | Reply

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

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