Question 1160668
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I'll do the first two problems to get you started.


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Problem 1


The first term is a = 3/5
The common ratio is r = -1/3


We multiply each term by -1/3 to get the next term
first term = -3/5
second term = (common ratio)*(first term) = (-1/3)*(3/5) = -1/5
third term = (common ratio)*(second term) = (-1/3)*(-1/5) = 1/15
etc etc


With r = -1/3 = -0.33 (approximately), we can see that -1 < r < 1 is true, which means that the infinite geometric sum converges to a finite number. That number is...
{{{S = a/(1-r)}}}


{{{S = (3/5)/(1-(-1/3))}}}


{{{S = (3/5)/(1+1/3)}}}


{{{S = (3/5)/(3/3+1/3)}}}


{{{S = (3/5)/(4/3)}}}


{{{S = (3/5)*(3/4)}}}


{{{S = 9/20}}}



Answer: <font color=red size=4> 9/20 </font>


With your calculator, note how 
<ul><li>3/5-1/5 = 0.4</li><li>3/5-1/5+1/15 = 0.46666666666667 (approximate)</li><li>3/5-1/5+1/15-1/45 = 0.44444444444444 (approximate)</li><li>3/5-1/5+1/15-1/45+1/135 = 0.45185185185186 (approximate)</li><li>3/5-1/5+1/15-1/45+1/135-1/405 = 0.44938271604939 (approximate)</li><li>3/5-1/5+1/15-1/45+1/135-1/405+1/1215 = 0.45020576131688 (approximate)</li></ul>Those partial sums are slowly approaching 9/20 = 0.45; they'll never actually get there because we can't actually reach infinity. The sums only get closer and closer.


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Problem 2


Same idea as problem 1. We have different values this time of course. In this problem, 
a = 1 is the first term
r = 1/4 is the common ratio


We can determine the common ratio by picking any term but the first one, then dividing it over its previous term


examples:
(second term)/(first term) = (1/4) divided by (1) = 1/4
(third term)/(second term) = (1/16) divided by (1/4) = (1/16)*(4/1) = 4/16 = 1/4
These are two of many ways to see why the common ratio is r = 1/4.


The infinite sum converges to a finite value because r = 1/4 = 0.25 makes -1 < r < 1 true. 


Side note: We can condense <font color=blue>-1 < r < 1</font> into the absolute value inequality <font color=blue> |r| < 1 </font>. This says r is within 1 unit of zero, or that the distance from r to 0 is less than 1.


Anyways, back to the problem. We plug a = 1 and r = 1/4 into the formula used in the prior question.


{{{S = a/(1-r)}}}


{{{S = (1)/(1-1/4)}}}


{{{S = 1/(4/4-1/4)}}}


{{{S = 1/(3/4)}}}


{{{S = 1*(4/3)}}}


{{{S = 4/3}}}


Answer: <font color=red size=4> 4/3 </font>


Partial check or partial confirmation we have the right answer:
<ul><li>1+1/4 = 1.25</li><li>1+1/4+1/16 = 1.3125</li><li>1+1/4+1/16+1/64 = 1.328125</li><li>1+1/4+1/16+1/64+1/256 = 1.33203125</li><li>1+1/4+1/16+1/64+1/256+1/1024 = 1.3330078125</li><li>1+1/4+1/16+1/64+1/256+1/1024+1/4096 = 1.333251953125</li></ul>The partial sums are gradually approaching 4/3 = 1.33333...


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