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Do it in segments remember that there are 12 months in a year. n=2
I tried R, refular deposits (payment) = 150, n, number of deposits or payment = 3, and i, interest rate per compounding period: 0.03765/2 = 0.01875. I got $458. 50 in the end using this formula: A = R [(1 + i)^n - 1 / i. However, the correct answers are: a) $617.09 and b) $17.09
n = 2 because it is compounded twice a year so every six months 3.75% of the money present in the account is added to the original amount invested.
ohh.. so what's the function of the 'he made 3 more deposits of $150.'?
We know the original amount was compounded once before he made his second investment, then compounded again when he made his third investment than compounded again when he made his fourth investment
yup, he did.
To answer b) The first compound can be expressed as (0.0375)*150 + 150 The second compound he added 150 so for the second compound (first year we have the expression) ((((0.0375)*150 + 150) + 150)0.03 + ((((0.0375)*150 + 150) + 150)
I would probably just try answering this manually :L
ohhh, okay. So do I still use the formula or.. no?
well you could use it for every compound but make sure you change the principle accordingly
I'm sure there is a way to modify the equation so that ever time there is a compound 150 is added plus the interest
ohh.. okay. ohhhh.. like this: R + R(1 + i)^1 + R(1 + i)^2 + R(1+i)^3 +... + R (1+i)^n-1? And calculate the sum of the series: which is basically the formula?
oh, that one is for compounding interests isn't it?
yeah but I was wrong the formula has to take in account the added 150 after each compound