Question 1001423: Please help me solve the problem
Let Sn(n>=1) be a sequence of sets defined by
S1 = {0}
S2 ={3/2,5/2}
S3={8/3,11/3,14/3}
S4={15/4,19/4,23/4,27/4},.......... then
a) third element in S20 is 439/20
b)third element in S20 is 431/20
c)sum of element in S20 is 589
d) sum of the elements in S20 is 609
[Sn denotes S subscript n eg. S1 denotes S subscript 1]
Which of the above options(a,b,c,d) are correct? {at least two of them must be correct}
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! 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
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