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

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- anonymous

@Michele_Laino can you help again please

- Michele_Laino

please wait a moment

- 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

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

- Michele_Laino

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

- 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

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

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

- Michele_Laino

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

- anonymous

I'm confused and a bit discouraged

- Michele_Laino

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

- anonymous

How would I explain this using steps

- 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

you can explain, using my replies above

- anonymous

oh ok

- 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

formula*

- anonymous

ok hold on

- Michele_Laino

ok! please wait...

- anonymous

ok i got it now

- 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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