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I came across a problem as follows - A bank pays interest @6% per annum compunded continuously. If a person places 1000 in bank each year (assuming in the end of each year), how much will be in his account after 3 years?

Now we have 2 methods to solve this problem, the first one is to use sum to a gp formula, so the FV will be FV = 1000 * (e^0.18 - 1) / (e^0.06 - 1) which gives me 3189.33 which is what I also get on using various online calculators for annuities.

However, the book used a different approach. It integrated the function 1000*e^0.06 from 0 to 3 which returns 3286.67.

Although, I can understand why it returns a different answer, however does it makes sense to integrate practically? Shouldn't we simply use sum to a gp formula to arrive at 3189.33?

Any help is highly appreciated.

Thank you

Best Sourabh

Sourabh Gupta
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1 Answers1

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Its about truth in sales. An integrated formula gives a true 6% compounded continuously over 3 years. Your other method falls short. I find it unusual for this particular scenario that there is no money in the account for the first year. The results of these calculations only accumulate 2 years worth of interest.

$\frac{dP}{dt} = .06P + 1000$

when $t = 0, P = 0$

eventually comes to:

$P(t) = 16666.67e^{.06t} - 16666.67$

P(3) = 3286.96

Phil H
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