Question 1001423
S1 has 1 element
S2 has 2 elements
S3 has 3 elements
...
...
S20 has 20 elements
Sn has n elements



The denominator of each term in S2 is 2
The denominator of each term in S3 is 3
The denominator of each term in S4 is 4
...
...
The denominator of each term in S20 is 20
The denominator of each term in Sn is n


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The numerators are a bit trickier to figure out. But notice how 


S2 has numerators 3,5
so we have +2 each time


S3 has numerators 8,11,14
we're adding 3 each time


S4 has numerators 15,19,23,27
we're adding 4 each time


so Sn will have us adding n each time
The question is: where do we start?


S2 starts with 3
S3 starts with 8
S4 starts with 15


It turns out that the sequence 3,8,15 follows a quadratic pattern. In other words, a quadratic goes through the three points (2,3), (3,8) and (4,15)


I'm not going to go into too much detail about this (since it takes a very long time to do by hand), but you use a calculator to get this quadratic y = x^2 - 1


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So Sn has n elements. The first element would be(n^2-1)/n


The first element of S20 would be...


(n^2-1)/n = (20^2-1)/20 = 399/20


Now we add 20 two times to get


399+20+20 = 439


So the third element of S20 is 439/20


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We start with 399 in the numerator and we have the sequence


399, 419, 439, 479, ...


we have 20 of these terms being added up. This can be done quickly with an arithmetic series formula


sum of first n terms = n*(a1 + an)/2
sum of first n terms = n*(a1 + a1 + d(n-1))/2
sum of first n terms = n*(2*a1 + d(n-1))/2
sum of first 20 terms = 20*(2*399 + 20(20-1))/2
sum of first 20 terms = 11,780


The sum of the terms in the numerator is 11,780


The denominator stays as 20


11,780/20 = 589



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Final Answers:

a) third element in S20 is 439/20
c) sum of element in S20 is 589