consider the kth product
(k³  1) / (k³ + 1)
factor the numerator:
(k1)(k²+k+1)
consider the k+1th product:
((k+1)³  1) / ((k+1)³ + 1)
expand the denominator:
k³ + 3k² + 3k + 2
and factor: (k+2)(k² + k + 1)
So the quadratic term in the kth numerator cancels the quadratic term in the k+1th demoninator, leaving a product (k1)*k*(k+1)... in the numerator, and a product (k+2)(k+3)... in the denominator, AND the very first denominator which has no prior numerator to cancel.
All of the (k+n) terms in the denominator cancel corresponding terms in the numerator, leaving just three terms in the numerator without matches: (k1)*k*(k+1), and the very first denominator is still left. That means the product overall is equal to:
(K³  K) / (K³ + 1) where K is the very first value of k: 2 for this problem.
Substituting, the product is (82) / (9) = 6/9 = 2/3

Posted by Paul
on 20130624 18:07:53 