Suppose that P dollars is invested in a savings account at interest rate I, compounded semiannually, for one year. The amount A in the account after one year is given by A = P(1 + i/2)^2
I got help with a problem similar to this earlier, but I still do not understand.
Thank you! :-)
interest rate is i per year
so it is i/2 for half a year
every half hear you multiply by (1+i/2)
so after 1 year, two half years, you multiply by (1+/2)(1+i/2) = (1+i/2)^2
say i = 20% = .20
that is 10% per half year or .10 added every half year
so after a half year we have P + .10 P = P(1.10)
after a whole year, multiply by (1.1 again
P (1.10)^2 = 1.21 P
Sure! I'd be happy to help you understand.
In this problem, you have an amount of money (P) that is invested in a savings account with an interest rate (I). The interest is compounded semiannually, which means it is calculated twice a year.
The formula A = P(1 + i/2)^2 represents the amount (A) in the account after one year. Let's break it down step by step:
1. First, we need to find the interest rate for each compounding period. Since the interest is compounded semiannually (twice a year), we divide the annual interest rate (I) by 2: i = I/2.
2. Next, we calculate the value of 1 + i/2. This represents the growth factor for each compounding period. For example, if the interest rate is 4%, then i = 0.04/2 = 0.02. So, 1 + i/2 = 1 + 0.02/2 = 1.01.
3. Now, we raise (1 + i/2) to the power of 2, which represents the number of compounding periods in one year. This gives us (1 + i/2)^2.
4. Finally, we multiply the initial amount (P) by the growth factor (1 + i/2)^2 to calculate the final amount (A) in the account after one year.
Here's an example to illustrate the calculation:
Suppose you invest $1000 in a savings account with an interest rate of 4%, compounded semiannually. Using the formula:
A = P(1 + i/2)^2
A = $1000(1 + 0.02)^2
A = $1000(1.01)^2
A = $1000(1.0201)
A ≈ $1020.10
So, after one year, you would have approximately $1020.10 in the account.
I hope this explanation helps you understand the problem better! Let me know if you have any further questions.