loan analysis program (Proof)

[leung at imsvax.UUCP]

Posted Feb 2 stevens at hsi.UUCP in net.source has a program for
loan analysis.  There is no analytic proof of the formula he used.
The following is a proof of the formula.

The following lemma will be useful.

Lemma.  

      n
      _
      \         i          n+1
1 + a /  (1 + a)  = (1 + a)              (*)
      -
     i=0

Proof:

For i=0, lhs = 1 + a = rhs
Suppose for i=n, (*) holds
For i=n+1,
		     n+2
	lhs = (1 + a)
			     n+1
	    = (1 + a) (1 + a)

			     n
			     _
			     \ 	       i
	    = (1 + a) (1 + a /  (1 + a) )
                             -
                            i=0

		       n+1
			_
			\	  i
	    = 1 + a + a /  (1 + a)
			-
		       i=1

		   n+1
		    _
		    \
	    = 1 + a /  (1 + a) = rhs
	            -
		   i=0			 Q.E.D.

Back to the Loan Equation:

Theorem:

 	       n
     AI (1 + I)
P = ------------
           n
    (1 + I)  - 1

where P = Monthly Payment
      A = Loan Amount
      I = Monthly Interest Rate
          (Annual Interest Rate / 12)
      n = Number of Payments

Proof:

For n=1, the final balance is

	A(1 + I) - P = 0

For n=2, the final balance is

	(A(1 + I) - P)(1 + I) - P = 0
        
		2
	A(1 + I)  - P(1 + (1 + I)) = 0

For n=3, the final balance is

		 2
	(A(1 + I)  - P(1 + (1 + I)))(1 + I) - P = 0


		3                          2
	A(1 + I)  - P(1 + (1 + I) + (1 + I) ) = 0

For n=n, the final balance is

		     n-1
		      _
		n     \	        j
	A(1 + I)  - p /  (1 + I)  = 0
		      -
		     j=0

then,

		       n-1
			_
		 n	\         j
	AI(1 + I)  - pI /  (1 + I)  = 0
			-
		       j=0

Using the above Lemma,

		 n            n
	AI(1 + I)  - p((1 + I)  - 1) = 0

This completes the proof.