anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@Michele_Laino can you help again please
Michele_Laino
  • Michele_Laino
please wait a moment
Michele_Laino
  • Michele_Laino
the first payment is 150 the second payment is 150-1.3*150 the third payment is 150-1.3*(150-1.3*150) and so on...

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More answers

anonymous
  • anonymous
I think there should be a faster way because there is 15 payments
Michele_Laino
  • Michele_Laino
yes! I wrote those formulas in order to show the pattern
Michele_Laino
  • Michele_Laino
I call with a the quantity 150, namely a= 150, then I call with k the percentage, namely k=1.3
Michele_Laino
  • Michele_Laino
so I can write this first payment= a second payment =a-ka third payment= a-k(a-ka)=a-ka-k^2 a fourth payment= a-k(a-ka-k^2 a)= a-ka -k^2 a- k^3 a
Michele_Laino
  • Michele_Laino
n-th payment= a-ka-k^a-...-k^(n-1) a
Michele_Laino
  • Michele_Laino
oops..n-th payment= a-ka-k^2 a-...-k^(n-1) a
anonymous
  • anonymous
I'm confused and a bit discouraged
Michele_Laino
  • Michele_Laino
I have written, using formulas, the text of your nproblem
anonymous
  • anonymous
How would I explain this using steps
Michele_Laino
  • Michele_Laino
here is the sigma-notation for n-th payment p(n): \[\Large p\left( n \right) = a - a\sum\limits_{i = 1}^{n - 1} {{k^i}} \]
Michele_Laino
  • Michele_Laino
you can explain, using my replies above
anonymous
  • anonymous
oh ok
Michele_Laino
  • Michele_Laino
my 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
  • Michele_Laino
formula*
anonymous
  • anonymous
ok hold on
Michele_Laino
  • Michele_Laino
ok! please wait...
anonymous
  • anonymous
ok i got it now
Michele_Laino
  • Michele_Laino
now 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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