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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.

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  1. anonymous
    • one year ago
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    @Michele_Laino can you help again please

  2. Michele_Laino
    • one year ago
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    please wait a moment

  3. Michele_Laino
    • one year ago
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    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...

  4. anonymous
    • one year ago
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    I think there should be a faster way because there is 15 payments

  5. Michele_Laino
    • one year ago
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    yes! I wrote those formulas in order to show the pattern

  6. Michele_Laino
    • one year ago
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    I call with a the quantity 150, namely a= 150, then I call with k the percentage, namely k=1.3

  7. Michele_Laino
    • one year ago
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    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

  8. Michele_Laino
    • one year ago
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    n-th payment= a-ka-k^a-...-k^(n-1) a

  9. Michele_Laino
    • one year ago
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    oops..n-th payment= a-ka-k^2 a-...-k^(n-1) a

  10. anonymous
    • one year ago
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    I'm confused and a bit discouraged

  11. Michele_Laino
    • one year ago
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    I have written, using formulas, the text of your nproblem

  12. anonymous
    • one year ago
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    How would I explain this using steps

  13. Michele_Laino
    • one year ago
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    here is the sigma-notation for n-th payment p(n): \[\Large p\left( n \right) = a - a\sum\limits_{i = 1}^{n - 1} {{k^i}} \]

  14. Michele_Laino
    • one year ago
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    you can explain, using my replies above

  15. anonymous
    • one year ago
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    oh ok

  16. Michele_Laino
    • one year ago
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    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

  17. Michele_Laino
    • one year ago
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    formula*

  18. anonymous
    • one year ago
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    ok hold on

  19. Michele_Laino
    • one year ago
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    ok! please wait...

  20. anonymous
    • one year ago
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    ok i got it now

  21. Michele_Laino
    • one year ago
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    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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