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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2 Responses to “MMM #27: We have a winner!”
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?
By IBY on Apr 6, 2009 | Reply
Wait, how do you get from the third step to the penultimate step?