Question

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.

Answer 1

@Michele_Laino can you help again please

Answer 2

please wait a moment

Answer 3

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...

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

I'm confused and a bit discouraged

Answer 11

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

Answer 12

How would I explain this using steps

Answer 13

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

Answer 14

you can explain, using my replies above

Answer 15

oh ok

Answer 16

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

Answer 17

formula*

Answer 18

ok hold on

Answer 19

ok! please wait...

Answer 20

ok i got it now

Answer 21

now we have to compute this sum:
S=p(1)+p(2)+...+p(15)
so we can write this:
\[\Large \begin{gathered}
S = p\left( 1 \right) + p\left( 2 \right) + p\left( 3 \right) + ... + p\left( {15} \right) = \hfill \\
= a + \sum\limits_{j = 2}^{15} {p\left( j \right)} = a + \sum\limits_{j = 2}^{15} {\left( {a - a\sum\limits_{i = 1}^{j - 1} {{k^i}} } \right)} \hfill \\
\end{gathered} \]