Amortization Table

Let $B_n$ be the balance you owe to the bank after $n$ payments and $M$ be the payments. Then
\[B_{n+1}=B_n+\frac{4}{100}\times \frac{1}{24} B_n-M=(1+\frac{1}{600})B_n-M.\]
So 
\[B_{n+1}=\alpha B_n-M, \]
where $\alpha=\frac{601}{600}.$ We have
\[B_{n+1}=\alpha B_n-M= \alpha (\alpha B_{n-1}-M)-M=\alpha^2 B_{n-1}-\alpha M-M\]
\[=\alpha^2 (\alpha B_{n-2}-M)-\alpha M-M=\alpha^3 B_{n-2}-\alpha^2 M-\alpha M-M.\]
Repeating this process we get that
\[B_{n+1}=\alpha^{n+1}B_0-(\alpha^n +\alpha^{n-1}+ \dots +\alpha+1)M=\alpha^{n+1}B_0-\frac{\alpha^{n+1}-1}{\alpha -1}M.\]
Hence we have the important formula 
\[B_{n+1}=\alpha^{n+1}B_0-\frac{\alpha^{n+1}-1}{\alpha -1}M.\]
In the given example we have $B_0=683,000$, and $M=1,300$. So 
\[B_{n+1}=(\frac{601}{600})^{n+1}683000-\frac{(\frac{601}{600})^{n+1}-1}{\frac{601}{600} -1}1300.\]
So
\[B_{n+1}=(\frac{601}{600})^{n+1}683000-780000[(\frac{601}{600})^{n+1}-1].\]
Or 
\[B_{n}=(\frac{601}{600})^{n}683000-780000[(\frac{601}{600})^{n}-1]\]
which simplifies to 
\[B_n=780000-97000(\frac{601}{600})^{n}.\]

For the first payment $\$1300$, the interest portion is
\[683000*\frac{4}{100}*\frac{1}{24}=\frac{683000}{600}=\$1138.33.\]
The principle portion is $1300-1138.33=\$161.67.$

The balance after the first payment is
\[683000-161.67=\$ 682838.33.\]

Next we compute the total number of payments $n$: 
\[B_{n}=780000-97000(\frac{601}{600})^{n}=0\]
\[\Rightarrow (\frac{601}{600})^{n}=\frac{780000}{97000}=8.04123\]
\[\Rightarrow n \ln \frac{601}{600}=\ln 8.04123  \Rightarrow n=\frac{\ln 8.04123}{ln(\frac{601}{600})}=1251.79\approx 1252.\]
Aftere $1251$ payments, the balance is 
\[B_{1251}=780000-97000(\frac{601}{600})^{1251}=1027.76.\]
So for the last payment the interst portion is
\[1027.76 \times \frac{4}{100}*\frac{1}{24}=\$1.71.\]
In the last payment the blnce is fully paid so the principle portion is
\[\$1027.76.\]
The last payment amount is
\[1027.76+1.71=\$1029.47.\]
The balance after the last payment is $\$0.00$.