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Could someone help me solve this... I already solved parts a) and b) ( the work is below) but i need to solve part c) but im not really sure how to do that.
1.
a)
Values of r
1. r=2
2.r=3
3.r=4
4.r=5
5.r=6
Sum
10 Terms 50 Terms 100 Terms
1. 2046 2.25 x 10^15 2.53 x 10^30
2. 59048 7.17 x 10^23 5.15 x 10^47
2. 699050 8.45 x 10^29 1.07 x 10^60
4. 4882812 4.44 x 10^34 3.94 x 10^69
5. 24186470 3.23 x 10^38 2.61 x 10^77
Work
1. Sn=2(1-2^10 )/(1-2) = 2046
Sn=2(1-2^50 )/(1-2) = 2.25 x 10^15
Sn=2(1-2^100 )/(1-2) = 2.53 x 10^30
2. Sn=2(1-3^10 )/(1-3) = 59048
Sn=2(1-3^50 )/(1-3) = 7.17 x 10^23
Sn=2(1-3^100 )/(1-3) = 5.15 x 10^47
3. Sn=2(1-4^10 )/(1-4) = 699050
Sn=2(1-4^50 )/(1-4) = 8.45 x 10^29
Sn=2(1-4^100 )/(1-4) = 1.07 x 10^60
4. Sn=2(1-5^10 )/(1-5) = 4882812
Sn=2(1-5^50 )/(1-5) = 4.44 x 10^34
Sn=2(1-5^100 )/(1-5) =3.94 x 10^69
5. Sn=2(1-6^10 )/(1-6) = 24186470
Sn=2(1-6^50 )/(1-6) = 3.23 x 10^38
Sn=2(1-6^100 )/(1-6)= 2.61 x 10^77
b)
Values of r
1. r=.9
2.r=.8
3.r=.7
4.r=.6
5.r=.5
Sum
10 Terms 50 Terms 100 Terms
1. 13.03 19.89 19.99
2. 8.92 9.999857275 9.999999998
3. 6.4783498 6.6666665 6.666666667
4. 4.969766912 5 5
5. 3.99609375 4 4
Work
1. Sn=2(1-〖.9〗^10 )/(1-.9) = 13.03
Sn=2(1-〖.9〗^50 )/(1-.9) = 19.89
Sn=2(1-〖.9〗^100 )/(1-.9) =19.99
2. Sn=2(1-〖.8〗^10 )/(1-.8) = 8.9262581
Sn=2(1-.8)/(1-.8) = 9.999857275
Sn=2(1-〖.8〗^100 )/(1-.8) = 9.999999998
3. Sn=2(1-〖.7〗^10 )/(1-.7) = 6.4783498
Sn=2(1-.7)/(1-.7) = 6.6666665
Sn=2(1-〖.7〗^100 )/(1-.7) = 6.666666667
4. Sn=2(1-〖.6〗^10 )/(1-.6) = 4.969766912
Sn=2(1-〖.6〗^50 )/(1-.6) = 5
Sn=2(1-〖.6〗^100 )/(1-.6) =5
5. Sn=2(1-〖.5〗^10 )/(1-.5) = 3.99609375
Sn=2(1-〖.5〗^50 )/(1-.5) = 4
Sn=2(1-〖.5〗^100 )/(1-.5)= 4
a) Chose 5 different values for r such that |r|> 1. For each of these geometric series, let the first term g1 = 2. Determine the sum of each of these geometric series for 10 terms, 50 terms, and 100 terms. Organize your data into a chart. (10 points)
b) Chose 5 different values for r such that
|r| <1 For each of these geometric series, let the first term g1 = 2. Determine the sum of each of these geometric series for 10 terms, 50 terms, and 100 terms. Organize your data into a chart. (10 points)
c) The word converge means to approach or draw near to a particular value. For
example, as x gets very large, converges to zero. The word diverge means
does not converge. For example as x gets very large, x^2 diverges; as x gets large, x^2 does not approach any particular value. Make a conjecture about the conditions under which a geometric series will converge. Test your conjecture using g1 (not equals) 2 and some of the values of r. Think about the formulas for geometric series. Write a mathematical argument to support your conjecture. (30 points)
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From Part b) you might begin to realise that if |r|<1, the terms will converge to zero eventually. You could use this as your conjecture for part c).
When testing your conjecture, just experiment around abit. See how the terms behave for different values of r.
The formula for a geometric series is
What do you think will happen when
and |r|<1?Offline
the sum would keep getting smaller and smaller....right?
technically it would never hit zero, because n is infinite, but it would be pretty darn close to zero
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what mathematical arguement would i be able to use to support my conjecture
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If |r|>1, you get a divergent series with every term greater than the preceding term.
In such cases, the sum of n terms as n tends to infinity cannot be determined. The sum tends to infinity as n tends to infinity.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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isnt the formula for a sum of a geometric series Sn= a1(1-r^n)/1-r
because yours is a little different
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so would i be able to use this as a mathematical arguement to support conjecture?
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isnt the formula for a sum of a geometric series Sn= a1(1-r^n)/1-r
This formula is applicable when |r|<1.
When |r|>1, the formula given in the earlier post would hold good.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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Technially both can be used in all cases, it's just that which the easier one is varies depending on |r|.
When |r|<1, sachik's formula has positive numerator and denominator, making it slightly easier than the other one. When |r|>1, the reverse is true.
Why did the vector cross the road?
It wanted to be normal.
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