January 200,000 *1.003375%-x
February (200,000 *1.003375-x) *1.003375-x = 200,000 *1.003375 2-(1.003377)
The third month [200,000 *1.003375 2-(1.003375+1) x] *1.003375-x
=200,000* 1.003375^3-( 1.003375^2+ 1.003375+ 1)x
......
Nth month 200,000 *1.003375n-[1.003375 (n-1)+1.003375 (n-2)] ...
an=200,000* 1.003375^n-[ 1.003375^(n- 1)+ 1.003375^(n-2)+......+ 1.003375^ 1+ 1)x
=200,000* 1.003375^n-[ 1+ 1.003375^ 1+......+ 1.003375^(n-2)+ 1.003375^(n- 1)]x
=200,000* 1.003375^n-[( 1.003375^n- 1)/( 1.003375- 1)]x
=200,000* 1.003375^n-( 1.003375^n- 1)/(0.003375）x
Pay off in ten years, that is, the balance will be zero after 120 months.
200,000* 1.003375^ 120-( 1.003375^ 120- 1)/(0.003375）x = 0
x=200,000* 1.003375^ 120/（ 1.003375^ 120- 1)/(0.003375）
=200,000* 1.003375^ 120 * 0.003375 /( 1.003375^ 120- 1)
You see if this is the case, and I won't count as a result, mainly the method.
I have checked, but I can't rule out mistakes. Please advise.