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JH
Numerade Educator

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Problem 66 Medium Difficulty

If you deposit $ \$ $100 at the end of every month into an account that pays $ 3 \% $ interest per year compounded monthly, the amount of interest accumulated after $ n $ months is given by the sequence
$ I_n = 100 \left( \frac {1.0025^n - 1}{0.0025} - n\right) $
(a) Find the first six terms of the sequence.
(b) How much interest will you have earned after two years?

Answer

(a) $$
I_{n}=100\left(\frac{1.0025^{n}-1}{0.0025}-n\right)
$$
(b) $\$ 70.28$

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Video Transcript

so we deposit a hundred dollars at the end of every month into an account that pays three percent interest per year. Compounded monthly. That's a key word here monthly. Then the amount of interest accumulated after and months is given by the following part. A. Let's find the first six terms. So that's just eye. One eye, too upto I six. Now using a calculator here in Wolfram, I have plotted out the first six terms here, going a few decimal places. So, of course, in our problem, it will make sense to round off to to places after the decimal, cause it's money so you can pause the screen. Recorded these values easily Your first six terms. I want through I six now for Part B. How much interest will you have earned after two years? Well, noticed that two years equals twenty four months, and it's important here that we're using months because this model here, this this formula is in terms of months, not years. So we use and equals twenty four. So really, in part B. What they're asking for is the value of I twenty four. So all you do plug in twenty four friend and then goto a calculator and go ahead and plug this in if you want an approximation and then minus end, which is twenty four. So that's the exact answer. And if you want to round off because it's this is a dollar amount lets this will be seventy point twenty eight. So seventy dollars in twenty eight cents after two years, and that's our answer for Part B.