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The question states: $\$500$ deposited @ time $= 0$, and then $\$1000$ deposited @ time $=3$ for a total of $12$ years. I need to find the value @ time = $12$. Discount rate = $7.5\% $

So, my equation is -> $$500\times (1.075)^{12}+1000\times (1.075)^9 = \$3108.128$$

In this case the expected answer is: $\$3291.38$ What am I going wrong ?

UPDATE: I figured it out.

$$500\times 1/(1-0.075)^{12}+1000\times 1/(1-0.075)^9 = \$3291.38$$

ATS
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  • I apologize, I will update that. – ATS Sep 16 '21 at 23:37
  • Also: don't use dollar signs unless you know how to use them in the formatting system. As you see, they don't have the effect you intend. – lulu Sep 16 '21 at 23:38
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    I see nothing wrong with your methodology. I experimented with various values close to $1.075$ and could not reverse-engineer the supposed answer. I also experimented with continuous interest given by $e^{it} ~: ~e \approx 2.71828$ and still could not reverse-engineer the supposed answer. – user2661923 Sep 16 '21 at 23:48
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    Finally, I doubt that the issue is simple versus compound interest, for two reasons: [1] Would be very unusual (and arguably pointless) mathematical exercise. [2] Then, the supposed answer would be lower, not higher. – user2661923 Sep 16 '21 at 23:50
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    @ATS, regarding your update, surely you mean $0.925$ instead of $1.075$. – Rushy Sep 17 '21 at 00:21

1 Answers1

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I write some details, which explains why your equation is right. The discount factor is

$$\frac{1}{1+i}=1-d,$$

where $d$ is the discount rate and $i$ is the interest rate. Keep that relation in mind. Now you want to compound the payments. For this purpose you take the reciprocal.

$$1+i=\frac1{1-d}=\frac1{1-0.075}=\frac1{0.925}$$

This is the factor for compounding. I hope it clarifies some things.

callculus42
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