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 April 3rd, 2014, 04:30 PM #1 Newbie   Joined: Oct 2011 Posts: 3 Thanks: 0 Finance & Compounding Interest It's been years since I studied math. I've forgotten the basics. I've come across this question, and I can't solve it (at least I'm not getting the same solution as a book). Here's the problem: How much money do you have to deposit into an account monthly, for 30 years, at an interest rate of 12%, compounded monthly, in order to have $1.5 million at the end of the 30 years? Here's what I have: y = ca^x, where c is a constant, a is the growth factor, and x is the number of months c E (1.12)^n (i = 1, n = 360) Therefore, 1.5x10^6 = c[1.12^360 + 1.12^359 + 1.12^358 + ... + 1.12] Solving for c (after computing the summation with a calculator) yields a very small number. I'm not sure my Sigma formula is correct, and I don't know how to calculate it by hand. I'd like to know how to manually do it. I totally forgot how to compute this like a human.  April 3rd, 2014, 06:45 PM #2 Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,163 Thanks: 472 Math Focus: Calculus/ODEs I would model this using a linear difference equation. Let$A_t$be the amount in the account at time$t$measured in months and$D$be the monthly deposit. We will let the clock begin at the moment the first deposit is made. Hence, our difference equation is:$\displaystyle A_{t+1}-1.01A_{t}=D$The homogeneous solution is:$\displaystyle h_t=c_1(1.01)^t=c_1\left(\frac{101}{100} \right)^t$and the particular solution is:$\displaystyle p_t=B$Using the method of undetermined coefficients, we find:$\displaystyle B=-100D$Thus, by the principle of superposition, the general solution is:$\displaystyle A_t=c_1\left(\frac{101}{100} \right)^t-100D$Using the initial value$A_0=D$, we may determine the value of the parameter:$\displaystyle A_0=c_1-100D=D\implies c_1=101D$And so the solution satisfying the model is:$\displaystyle A_t=D\left(101\left(\frac{101}{100} \right)^t-100 \right)$Now, to find find the value of$D$such that:$\displaystyle A_{360}=1500000$we may state:$\displaystyle D\left(101\left(\frac{101}{100} \right)^{360}-100 \right)=1500000$Solving for$D$, we find:$\displaystyle D=\frac{1500000}{101\left(\frac{101}{100} \right)^{360}-100}\approx424.82$April 3rd, 2014, 11:11 PM #3 Newbie Joined: Oct 2011 Posts: 3 Thanks: 0 Quote:  Originally Posted by MarkFL I would model this using a linear difference equation. Let$A_t$be the amount in the account at time$t$measured in months and$D$be the monthly deposit. We will let the clock begin at the moment the first deposit is made. Hence, our difference equation is:$\displaystyle A_{t+1}-1.01A_{t}=D$The homogeneous solution is:$\displaystyle h_t=c_1(1.01)^t=c_1\left(\frac{101}{100} \right)^t$and the particular solution is:$\displaystyle p_t=B$Using the method of undetermined coefficients, we find:$\displaystyle B=-100D$Thus, by the principle of superposition, the general solution is:$\displaystyle A_t=c_1\left(\frac{101}{100} \right)^t-100D$Using the initial value$A_0=D$, we may determine the value of the parameter:$\displaystyle A_0=c_1-100D=D\implies c_1=101D$And so the solution satisfying the model is:$\displaystyle A_t=D\left(101\left(\frac{101}{100} \right)^t-100 \right)$Now, to find find the value of$D$such that:$\displaystyle A_{360}=1500000$we may state:$\displaystyle D\left(101\left(\frac{101}{100} \right)^{360}-100 \right)=1500000$Solving for$D$, we find:$\displaystyle D=\frac{1500000}{101\left(\frac{101}{100} \right)^{360}-100}\approx424.82$That's pretty neat. The book's answer was$429, but that could be an error (they don't show their work).

I am not familiar with your method (despite what I read on Google a few minutes ago).

As for my Summation Notation Equation, was my equation incorrect? That is:

cE1.12^n (i=1, n=360)

 April 3rd, 2014, 11:31 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,163 Thanks: 472 Math Focus: Calculus/ODEs It could be that they intended an effective rate of 12% which means we would get the following monthly interest rate instead: $\displaystyle r=0.94888\overline{3}\%$ And this would change the deposit amount slightly...but I don't want to go through the derivation again. You need to use a monthly rate instead of the annual rate in your summation, that's why you are getting a small value.

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