Expert: Josh - 1/10/2005

Question
I am making a recurring deposit of USD 10,000 every year for 40 years.  The maturity amount after 40  years is USD 2,260,000.  The recurring deposit is earning interest compounded yearly.  How do I work out the rate of interest?

Using excel program, I could know that the rate of interest is 7.2106%.  But if I have to calculate the same manually, what would be the formula?

Answer
Dear Mr Bhatia,

My derivation leads to the following formula,
A(N)=a[C(N+1,1)+C(N+1,2)*r+C(N+1,3)*r^2+...+C(N+1,N)*r^(N-1)+C(N+1,N+1)*r^N]
..............................[Equation 1]
where
A(N) denotes the amount at maturity when time period n=N,
a is your annual deposit (of USD 10,000),
r is the annual interest rate and
C(m,j) is defined as m!/[(m-j)!*j!].

Let me say that this is not an easy equation to solve without a computer, because you have a high-order polynomial in terms of the unknown "r".
The solution I found using a mathematical package is approximately 7.19%.

The mathematical symbol m!, is called the "m factorial". It is the product of integers from m to 1. Specifically, it is given by m*(m-1)*(m-2)*....*2*1.
Zero factorial (0!) equals to 1, by definition.

The terms C(m,j) = m!/[(m-j)!*j!] represent the coefficients of the polynomial in "r".
For example, C(N+1,3)=(N+1)!/[(N+1-3)!*3!]
=(N+1)*(N)*(N-1)*(N-2)*(N-3)...*2*1/[(N-2)*(N-3)...*2*1 * 3*2*1]
=(N+1)*(N)*(N-1)/[3*2*1] after simplification.

Note: The formula is based on observation of the pattern which emerges from the following derivation.
A(0)=a,
A(1)=a(1+r),
A(2)=(a(1+r)+a)(1+r)=a(2+r)(1+r)=a(2+3r+r^2)
A(3)=[A(2)+a](1+r)=a(3+3r+r^2)(1+r)=a(4+6r+4r^2+r^3)
A(4)=[A(3)+a](1+r)...and so forth.

Disclaimer: Equation 1 should only be used as a guide only. Seek professional advice from qualified financial advisors before making any investment decision.