Calculus with interest rates

Question

Suppose a person will start academic studies on a private university in exactly $3$ years. The academic studies takes $4$ years and in the beginning of each of these four years the person has to pay $8000$ tuition fees. How much money would the person need at least, if her money out aside has an interest rate of $5\%$ (using exp. interest payment)?

My idea was the following:

$$C(t) = C(0)(1+r)^t,$$ with $C(t)=32000$, since in the end the person will need 32000. Applying above equation I get

$$32000 = C(0)(1+0.05)^7$$ and therefore C(0) = 22741.80.

Is this approach correct or does it have to be solved differently? Thanks for you help.

Answer 1

The capital at the beginning of year $t$ is: \begin{align} C(0) &= C \\ C(1) &= C (1 + r) \\ C(2) &= C (1 + r)^2 \\ C(3) &= C (1 + r)^3 - 8000 \\ C(4) &= (C (1 + r)^3 - 8000)(1 + r) - 8000 \\ C(5) &= ((C (1 + r)^3 - 8000)(1 + r) - 8000)(1 + r) - 8000 \\ C(6) &= (((C (1 + r)^3 - 8000)(1 + r) - 8000)(1 + r) - 8000) (1+r) - 8000 = 0 \end{align} Solving for $C$ in the last equation: $$ 0 = (((C (1 + r)^3 - 8000)(1 + r) - 8000)(1 + r) - 8000) (1+r) - 8000 \iff \\ (8000 / (1 + r)) = ((C (1 + r)^3 - 8000)(1 + r) - 8000)(1 + r) - 8000 \iff \\ ((8000 / (1 + r)) + 8000) / (1+r) = (C (1 + r)^3 - 8000)(1 + r) - 8000 \iff \\ ((((8000 / (1 + r)) + 8000) / (1+r)) + 8000) / (1+r) = C (1 + r)^3 - 8000 \iff \\ ((((((8000 / (1 + r)) + 8000) / (1+r)) + 8000) / (1+r)) + 8000) / (1+r)^3 = C $$ I told this my friend Ruby

irb(main):001:0> x1 = 8000.0 / 1.05
=> 7619.047619047618
irb(main):002:0> x2 = (x1 + 8000) / 1.05
=> 14875.283446712017
irb(main):003:0> x3 = (x2 + 8000) / 1.05
=> 21785.984234963827
irb(main):004:0> x4 = (x3 + 8000) / 1.05**3
=> 25730.253091427556
irb(main):005:0> C = x4
=> 25730.253091427556

and also tested this result with her help

irb(main):007:0> c1 = C * 1.05
=> 27016.765745998935
irb(main):008:0> c2 = C * 1.05**2
=> 28367.60403329888
irb(main):009:0> c3 = C * 1.05**3 - 8000
=> 21785.984234963827
irb(main):010:0> c4 = c3 * 1.05 - 8000
=> 14875.28344671202
irb(main):011:0> c5 = c4 * 1.05 - 8000
=> 7619.04761904762
irb(main):012:0> c6 = c5 * 1.05 - 8000
=> 1.8189894035458565e-12

which looks reasonable to me, neglecting rounding issues.