Compound Interest Millionaire

First of all, note that Barry's annualized gains remain constant throughout the life of his account, so his monthly gains are constant at a rate of $100 \left[ \left( 1.075 \right) ^ {1/12} - 1 \right] \approx 0.604491902429\%$ per month. We know that Barry adds $\$400.00$ to his investments every month, so if we let $M_k$ denote the expected value of the account at the beginning of the $k^{th}$ month of consecutive $\$400.00$ investments, observe that $M_{k+1} = M_k \cdot 1.00604491902429 + \$400.00, M_{k+2} = M_{k+1} \cdot 1.00604491902429 + \$400.00, M_{k+3} = M_{k+2} \cdot 1.00604491902429 + \$400.00$ , and so on. We therefore obtain, upon expansion of the recursively defined values, the following formula for the expected value of the account as a function of the number of months, $n$ , after the portfolio's current value, $M_k$ :

$\displaystyle{M_{k+n} = 1.00604491902429^n \cdot M_k + 400.00 \underbrace{ \left( 1.00604491902429 \times \left[ 1.00604491902429 \times \left[ \cdots \times \left[ 1.00604491902429 + 1 \right] + 1 \right] + \cdots \right]+ 1 \right)}_{\text{nested n-1 times.}}}$

So we see that the following equations hold:

$M_{k+n} = 1.00604491902429^n M_k + 400.00 \left( 1.00604491902429^{n-1} + 1.00604491902429^{n-2}+ \cdots + 1.00604491902429^2 + 1.00604491902429 + 1 \right)$

$\implies M_{k+n} = 1.00604491902429^n M_k + 400.00 \sum_{m=0}^{n-1} 1.00604491902429^m = 1.00604491902429^n M_k + 400.00 \left(\dfrac{1.00604491902429^n - 1}{1.00604491902429 - 1}\right).$

Letting $k$ be zero, we see that $M_n = 1.00604491902429^n \times 5000.00 + 400.00 \left( \dfrac{1.00604491902429^n - 1}{0.00604491902429} \right)$ .

Of course, we would like to find $n$ and not $M_n$ but it is not too difficult now to solve the equation for $n$ and see that

$n = \dfrac{ \log { \left[ \left( M_n \times 0.00604491902429 / 400.00 + 1 \right) / \left( 1 + 5000.00 \times 0.00604491902429 / 400.00 \right) \right] } } { \log{ \left( 0.00604491902429 \right) } }.$

With that, we see that for $M_n = \$1,000,000.00$ , we must have $n \approx 449.123$ months. So Barry's account balance will first exceed $\$1,000,000.00$ after $\lceil 449.123 \rceil = 450$ months.