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.

1. anonymous

@Michele_Laino can you help again please

2. Michele_Laino

3. 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...

4. anonymous

I think there should be a faster way because there is 15 payments

5. Michele_Laino

yes! I wrote those formulas in order to show the pattern

6. Michele_Laino

I call with a the quantity 150, namely a= 150, then I call with k the percentage, namely k=1.3

7. 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

8. Michele_Laino

n-th payment= a-ka-k^a-...-k^(n-1) a

9. Michele_Laino

oops..n-th payment= a-ka-k^2 a-...-k^(n-1) a

10. anonymous

I'm confused and a bit discouraged

11. Michele_Laino

I have written, using formulas, the text of your nproblem

12. anonymous

How would I explain this using steps

13. Michele_Laino

here is the sigma-notation for n-th payment p(n): $\Large p\left( n \right) = a - a\sum\limits_{i = 1}^{n - 1} {{k^i}}$

14. Michele_Laino

you can explain, using my replies above

15. anonymous

oh ok

16. 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

17. Michele_Laino

formula*

18. anonymous

ok hold on

19. Michele_Laino

20. anonymous

ok i got it now

21. 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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