# What is the formula for the difference between CI and SI?

if principal, time and rate are given how do i find the difference between Compound interest and Simple Interest?

P=12,000
n=1 and a  1/2 yrs.
R=10% per year


formulae that i know:

CI - SI for 2 years = P(R/100)^2
CI-SI for 3 years = P(R/100)^2 (R/100 + 3)


but none of these will work for 1 and a half years, so what formula do i use? or how do i use these formulae in this context?

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Welcome to Math.SE. I assume CI stands for Compound Interest and SI is Simple Interest? For future reference, it is generally good practice to define all your abbreviations and indicate what your variables stand for. Also, we'd prefer if you use other tags in addition to the (homework) tag. –  Willie Wong Jul 26 '12 at 16:01

Formula for simple interest:

$$P_{SI} = P \left(1 + \frac{nR}{100}\right)$$

Formula for compound interest:

$$P_{CI} = P \left( 1+\frac{R}{100} \right)^n$$

Therefore their difference is

$$P_{CI} - P_{SI} = P \left( \left(1+\frac{R}{100}\right)^n - \left(1+\frac{nR}{100}\right)\right)$$

If you substitute $n=2$ and $n=3$ into this formula, and expand out the brackets, you will get the formula you quoted in your question.

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alright, thanks –  Nirvan Jul 26 '12 at 16:14
The general compound interest formula says that after $n$ terms at a rate of $R$ percent per term is that the final principal is $P(1+\frac R{100})^n$. Often $R$ is quoted as an annual rate, but if you compound monthly you need to use $\frac R{12}$ per month and $n$ is the number of months. If you have partial terms, you need to specify what happens for a partial term. Maybe you get nothing for the last half year, maybe you get half the interest, or whatever. For simple interest, the final principal is $P(1+\frac {nR}{100})$, so the difference is just the difference of these: $P(1+\frac R{100})^n-P(1+\frac {nR}{100})$
The 2 year one is correct, which you can see by expanding the square (n=2) and the only term which survives is the R^2 term. The other is also correct if by $r$ you meant $R$, as it comes from $(1+a)^3-(1-3a)=(1-3a+3a^2+a^3)-(1-3a)=3a^2-a^3)$. My equation works for any $n$. You can see the effect is proportional to $(nR)^2$, so if that is very small you can ignore the difference. –  Ross Millikan Jul 26 '12 at 16:15