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anonymous
 one year ago
Larry has taken out a loan for college. He started paying off the loan with a first payment of $150. Each month he pays, he wants to pay back 1.3 times as the amount he paid the month before. Explain to Larry how to represent his first 15 payments in sigma notation. Then explain how to find the sum of his first 15 payments, using complete sentences. Explain why this series is convergent or divergent.
anonymous
 one year ago
Larry has taken out a loan for college. He started paying off the loan with a first payment of $150. Each month he pays, he wants to pay back 1.3 times as the amount he paid the month before. Explain to Larry how to represent his first 15 payments in sigma notation. Then explain how to find the sum of his first 15 payments, using complete sentences. Explain why this series is convergent or divergent.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino can you help again please

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1please wait a moment

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the first payment is 150 the second payment is 1501.3*150 the third payment is 1501.3*(1501.3*150) and so on...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think there should be a faster way because there is 15 payments

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! I wrote those formulas in order to show the pattern

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I call with a the quantity 150, namely a= 150, then I call with k the percentage, namely k=1.3

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so I can write this first payment= a second payment =aka third payment= ak(aka)=akak^2 a fourth payment= ak(akak^2 a)= aka k^2 a k^3 a

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1nth payment= akak^a...k^(n1) a

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1oops..nth payment= akak^2 a...k^(n1) a

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm confused and a bit discouraged

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I have written, using formulas, the text of your nproblem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How would I explain this using steps

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here is the sigmanotation for nth payment p(n): \[\Large p\left( n \right) = a  a\sum\limits_{i = 1}^{n  1} {{k^i}} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1you can explain, using my replies above

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1my for mula above, namely: \[\Large p\left( n \right) = a  a\sum\limits_{i = 1}^{n  1} {{k^i}} \] is valid for n>1 whereas for n=1, we have p(1)= a=150

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! please wait...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now we have to compute this sum: S=p(1)+p(2)+...+p(15) so we can write this: \[\Large \begin{gathered} S = p\left( 1 \right) + p\left( 2 \right) + p\left( 3 \right) + ... + p\left( {15} \right) = \hfill \\ = a + \sum\limits_{j = 2}^{15} {p\left( j \right)} = a + \sum\limits_{j = 2}^{15} {\left( {a  a\sum\limits_{i = 1}^{j  1} {{k^i}} } \right)} \hfill \\ \end{gathered} \]
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