This Questions seems daunting but let’s put our thinking in order.
Let’s suppose that for some (n) there exist the desired numbers. we also assume that S1 <S2 <S3<…<Sn
surely S1 > 1 else (1- 1/S1) = 0
so we have 2 ≤ S1 ≤ S2–1 ≤ S3–2≤…Sn -(n-1) , hence Si ≥i+1 for each i = 1,2…n. Therefore
51/2010 = 17/670 =
Shifu: Hello Po, it seems you’ve got a good challenge to solve… how do you plan to go about it?
Po: Master shifu, thanks but I guess there must be some tricks to solve it! I guess I should first get the expression in simplest form and then apply the…
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Shifu: How do we find the electric field?
Po: if it’s a point charge we use the formula KQ/r² otherwise we divide the charge into little pieces, find the electric field from each (as point charge), and add the electric fields (as vectors).
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