Ok guest, if you were a member then I would not identify you incorrectly. Yes this question is much more involved but the two looked like they came from the same person to me.
Your son will be going to college in 10 years and you are starting a fund for his education. He will need $25,000 at the beginning of each year for four years. The fund earns 5% annually, compounded monthly, and you plan to make monthly deposits, starting at the end of the current month. How much should you deposit each month to meet his educational expenses? It is assumed that the parents will continue to make the same deposits while their son continues his education for the 4-year term. Thanks for help. P.S. I know the answer but don't know how to arrive at it.
Lets say you invest $M at the end of every month for 1 year at 5% pa compounded monthly.
It will grow to
\(FV_{12}=M\left [\frac{1.0041\dot6^{12}-1}{0.0041\dot6}\right]=12.27885348M \quad dollars.\)
10 years is 120 months. How much will be there after 120 months.
\(FV_{120}=M\left [\frac{1.0041\dot6^{120}-1}{0.0041\dot6}\right]=155.282248M \quad dollars.\)
So, after 120 months thee will be 155.282248M - 500000 in the fund (that is after the first payment)
How much will be in the fund 12 months later?
\(T_{after\;second \;payment}=(155.282248M-50000)*1.00416^{12}+12.27885348M-50000\\ T_{after\;second \;payment}=155.282248*1.00416^{12}M-50000*1.00416^{12}+12.27885348M-50000\\ T_{after\;second \;payment}=(155.282248*1.00416^{12}+12.27885348)M-50000(1.00416^{12}+1)\\ \\~\\ T_{after\;third \;payment}=[(155.282248*1.00416^{12}+12.27885348)M-50000(1.00416^{12}+1)]*1.00416^{12}+12.27885348M-50000\\ T_{after\;third \;payment}=(155.282248*1.00416^{12}+12.27885348)*1.00416^{12}M-50000(1.00416^{12}+1)*1.00416^{12}+12.27885348M-50000\\ T_{after\;third \;payment}=[(155.282248*1.00416^{24}+12.27885348*1.00416^{12}]M-50000(1.00416^{24}+1.00416^{12})+12.27885348M-50000\\ T_{after\;third \;payment}=[(155.282248*1.00416^{24}+12.27885348*1.00416^{12}+12.27885348]M-50000(1.00416^{24}+1.00416^{12}+1)\\ T_{after\;third \;payment}=[(155.282248*1.00416^{24}+12.27885348*1.00416^{12}+12.27885348]M-50000(1.00416^{24}+1.00416^{12}+1)\\ Let Z=155.282248, \quad R=1.0041\dot6,\quad Y=12.27885348 \\ T_{after\;third \;payment}=[(Z*R^{24}+Y*R^{12}+Y]M-50000(R^{24}+R^{12}+1)\\ \text{following the pattern}\\ T_{after\;forth \;payment}=[(Z*R^{36}+YR^{24}+YR^{12}+Y]M-50000(R^{36}+R^{24}+R^{12}+1)\\ \)
\(Let Z=155.282248, \quad R=1.0041\dot6,\quad Y=12.27885348 \\ T_{after\;forth \;payment}=[(Z*R^{36}+YR^{24}+YR^{12}+Y]M-50000(R^{36}+R^{24}+R^{12}+1)\\\)
155.282248*1.0041666666666^36+12.27885348(1.0041666666666^24+1.0041666666666^12+1) = 219.1093482417387751879582937628068330956451487032115301294171720182
50000*(1.0041666666666666^36+1.0041666666666666^24+1.0041666666666666^12+1) = 215878.7732386761695511346
T (after four payments) = 219.1093482417M - 215878.7732386
but
T (after four payments) = 0
219.1093482417M - 215878.7732386=0
219.1093482417M = 215878.7732386
M=215878.7732386 / 219.1093482417
215878.7732386/219.1093482417 = 985.2558778115831659
I get $985 per month.
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Blast... I worked it out on $50000 every your for 4 years. If you change all the 50,000 to 25,000 you should get the right answer.
155.282248*1.0041666666666^36+12.27885348(1.0041666666666^24+1.0041666666666^12+1) = 219.1093482417387751879582937628068330956451487032115301294171720182
25000*(1.0041666666666666^36+1.0041666666666666^24+1.0041666666666666^12+1) = 107939.3866193380847755673
219.109348241738775M-107939.386619338=0
M=107939.386619338/219.109348241738775
107939.386619338/219.109348241738775 = 492.627938905877833739
So the answer is $493 per month. That sounds about right.
Is that at least close to the answer you were expecting?
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