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nth term question

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by Ian Stewart » Thu Feb 04, 2010 2:28 pm
By substitution we find that

t_3 = 3/5
t_4 = 4/6
t_5 = 5/7
...
t_51 = 51/53
t_52 = 52/54
t_53 = 53/55

so when we multiply these, the result is

(3/5)(4/6)(5/7)(6/8) ... (51/53)(52/54)(53/55)

Notice that almost all of the numerators will cancel with the denominator of the term two earlier (so the numerator in 5/7 cancels with the denominator of 3/5, for example); after canceling, all we have left is

(3*4)/(54*55) = 2/(9*55) = 2/495
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by neelimareddym » Thu Feb 04, 2010 10:20 pm
harsh.champ wrote:The question is in the attachment file.

The answer choices are
(1)2/495
(2)2/477
(3)12/55
(4)1/485
(5)1/2970
tn =n/(n+2)

=> t3=3/5
t4 = 4/6
t5=5/7
t6=6/8
...
...
...
t51 =51/53
t52=52/54
t53 = 53/55

Series = (3/5)*(4/6)*(5/7)*...*(51/53)*(52/54)*(53/55)
=(3*4)/(54*55)
=2/495
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by thephoenix » Thu Feb 04, 2010 10:38 pm
t3=3/5
t4=4/6
t5=5/6

hence the pattern is
3/5*4/6*5/7...................51/53*52/54*53/55
the first two numerator and the last two denominator will left and all others will get cancel out

hence
3*4/54*55=2/495
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