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by pemdas » Tue Jan 31, 2012 1:40 pm
if we follow all the way into algebra and find the formula for all n-s then we end up in something like 4(n!*n!/2 + n!*n!/3 + ... )/n!*n!

Instead I would use the shortcut and look into the first term which is 2 and the last term which is (4/100 - 4/101). It's clear that the answer cannot be 2 or 0 or even negative value as we have the positive value sequence and the first term is already 2. The only options are 3 and 4, so we narrowed the choices to "2 out of 5", better now ... and either D or E

To evaluate what is the answer I would propose, completing some more values here. We know by simple plugging into sequence function 4/(n^2+n) the 2nd term is 2/3 and the 3rd, 4th and 5th terms will be 1/3, 1/5, 2/15. Thus, (2/3 + 1/3 +1/5 +2/15) is (20+10+6+4)/30=4/3 OR greater than 1. We definitely cannot select answer D now as the resultant is more than 3, we have to mark choice E

correct answer e
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by neelgandham » Tue Jan 31, 2012 2:17 pm
an = (4/n)-(4/(n+1))

a1 = 4/1 - 4/2
a2 = 4/2 - 4/3
a3 = 4/3 - 4/4
.
.
.
.
a99 = 4/99 - 4/100
a100 = 4/100 - 4/101

a1 + a2 + ....a99 + a100 = 4/1 - 4/2 + 4/2 - 4/3 + 4/3 - 4/4.........+ 4/99 - 4/100 + 4/100 - 4/101
If you observe, except for the values at the extremes, i.e., 4/1 and -4/101, rest cancel out. So,
a1 + a2 + ....a99 + a100 = 4/1 - 4/101 = 4 - 4/100(approx) = ~ 3.96

Answer E
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