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Formula for calculating interest in Recurring deposit?

NickJr - 2006-08-09 06:32:19 - Mathematics

If I am making a recurring deposit of $XX.xx every year for YY years and the recurring deposit is earning interest compounded yearly, How do I work out the rate of interest? How to calculate the interest paid for the deposit(s)? What is the formula for calculating it?

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Best Answer:

Amount=Pn+(Pn(n+1)*r/2400) where P is the monthly deposit,n is the no of months,r is the annual rate of interest

Answers:

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Formula for calculating recurring deposit, Hope this helps.. 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. Note: Equation 1 should only be used as a guide only. Seek professional advice from qualified financial advisors before making any investment decision.

Amount=Pn+(Pn(n+1)*r/2400) where P is the monthly deposit,n is the no of months,r is the annual rate of interest

The exact formula for the total balance after Y years assuming an amount X was deposited at the beginning of each of those Y years is: A = X*(1+r)*((1+r)^Y-1)/r where r is the fixed annual interest rate. (The expressions given by Kae for specific Y values are correct but this is the general formula). The total amount deposited is XY, so the total interest paid is: I = A - XY = X*(1+r)*((1+r)^Y-1)/r - XY ***** To use a concrete example, say $1000 is deposited at the beginning of each year for 10 years at an interest rate of 5%. The ending balance and total interest paid are: A = 1000*(1.05)(1.05^10-1)/(.05) = $13,206.79 I = $13,206.79 - $10,000 = $3,206.79 ***** To see where the above formula comes from, note that the total present amount due to a deposit X made k years ago is: X*u^k where u = (1+r). So the total amount at the end of Y years is the sum of a geometric series: A = X*[u + u^2 + u^3 +...+u^Y] = X*[(u^(Y+1)-u)/(u-1)]

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