Posted by mike on Sunday, June 6, 2010 at 7:45pm.
(k(k-1))/2!*(1/k)^2 square 1/k and group into one single faction
= (k(k-1))/(2!*(k)^2) the k on top and the k on bottom cancel
= (k-1)/(2!*k) you can rewrite this as:
= (1/2!)*((k-1)/k) now multiply the whole thing by (1/k)/(1/k) {note: this is legal because it is the same as multiplying a number by 1}
= (1/2!)*((1/k)(k-1))/((1/k)*k) the bottom cancel out {1/k*k = 1} and just multiply the top out
= (1/2!)*(1-(1/k)) rewrite as:
= (1-(1/k))/2!
Fun problem!
What happened to the k in the 2!*k before you multiplied everything by (1/k)/(1/k)
(k(k-1))/2! * (1/k)^2
= (k²-k)/k² / 2!
= (k²/k² - k/k²) /2!
= (1 - 1/k)/2!
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