Question

What is the value of an investment of $6,835.69 at 3.88% interest compounded daily for 6 years?

Answer 1

I'm sorry, I'm not good with financial mathematics

Answer 2

1st year:
0.0388 x 6835.69 = $265.22 (2 decimal places)
6,835.69 + 265.22 = $7100.91

2nd year:
0.0388 x 7100.91 = $275.52 (2 decimal places)
7100.91 + 275.52 = $7376.43

3rd year:
0.0388 x 7376.43 = $286.21 (2 decimal places)
7376.43 + 286.21 = $7662.64

4th year:
0.0388 x 7662.64 = $297.31 (2 decimal places)
7662.64 + 297.31 = $7959.95

5th year:
0.0388 x 7959.95 = $308.85 (2 decimal places)
7959.95 + 308.85 = $8268.80

6th year:
0.0388 x 8268.80 = $320.83 (2 decimal places)
8268.80 + 320.83 = $8589.63

Does this make sense? It's 3.88% interest on the new amount each year.
I know this is closed, but I just thought I'd answer anyway haha. :)

Answer 3

Im not sure if its right, considering it says daily o.o
If not, I'm sorry xD But you might be able to work it out

Answer 4

Assuming the 3.88% is an annual rate, this is a pretty silly question

the *wrong* way to calculate a daily rate r from an annual rate R is to say \(r = {R \over 365}\) and so the answer to this question would be \( $6,835.69 (1+{3.88\% \over 365})^{6 \times 365} = $8,627.42 \)

the *correct* way is to say the daily \(r\) follows from the fact that \((1+r)^{365} = 1+ R\) so \(r=(1+R)^{{1\over 365}}-1\) but this means that the final amount \( A\) of initial investment \( A_o\) follows from \(A = A_o(1+r)^N = A_o((1+R)^{{1\over 365}})^N\) where \(N = 6 \times 365\)

which is just \(A = A_o(1+R)^6 \quad \quad [= $6,835.69(1+3.88 \%)^6 = $8,589.62 ]\)

ie the correct way, decompounding for a daily rate and then compounding back, is a waste of time and you may as well just look at it annually.