Problem 3
On January 1, 2010, Sergei signed a one-year Cable TV contract with Comcast, which specifies that he provide payments of $45 at the end of each month for 12 months, with the first payment beginning at the end of the month. If he had instead deposited these 12 monthly payments into a savings account at an annual effective interest rate of 4%, with interest paid at the end of every month, how much money would he have in the account by the end of the year?
Solution 3
The problem gives us an annual effective interest rate i=4%, but we need to know the monthly effective interest rate i(12)/12 because the payments occur every month. To find the monthly effective interest rate, we need to use the relation:
$latex (1+i)=(1+frac{i^{(m)}}{m})^{(m)}$
The left side of the equation tells us how much an account governed by an annual effective rate of interest i grows in one year. For instance, an account governed by an annual effective interest rate i will grow by (1+i)x(100%) each year. The right side of the equation tells us how much an account governed by an interest rate of i(m)/m per m-th of a year grows in one year. Using algebra to rearrange the equation, we have:
$latex displaystyle (1+i)^{frac{1}{m}}-1=frac{i^{(m)}}{m}$
By setting m=12, we can use the relationship to find the annual effective interest rate equivalent to the monthly effective rate:
$latex displaystyle (1+i)^{frac{1}{12}}-1=frac{i^{(12)}}{12}$
Setting the annual effective rate i=4%=.04 yields us:
$latex displaystyle (1+.04)^{frac{1}{12}}-1=frac{i^{(12)}}{12}approx 0.00327374$
Now that we have the monthly effective rate of interest, we can find the accumulated value of the twelve payments. Each of these payments is governed by the general accumulation formula, which yields us the total accumulated value of:
$latex displaystyle 45(1+frac{i^{(12)}}{12})^{11}+45(1+frac{i^{(12)}}{12})^{10}+dotsm+45(1+frac{i^{(12)}}{12})^1+45(1+frac{i^{(12)}}{12})^0$
Factoring out the 45 gives us:
$latex displaystyle 45[(1+frac{i^{(12)}}{12})^{11}+(1+frac{i^{(12)}}{12})^{10}+dotsm+(1+frac{i^{(12)}}{12})^1+(1+frac{i^{(12)}}{12})^0]$
We can solve this sum right now if we wanted, but the calculation is tedious if we do it now. However, there is a way to simplify the terms inside the square brackets. Notice that the sum inside the square brackets:
$latex displaystyle (1+frac{i^{(12)}}{12})^{11}+(1+frac{i^{(12)}}{12})^{10}+dotsm+(1+frac{i^{(12)}}{12})^1+(1+frac{i^{(12)}}{12})^0$
Is equal to a geometric sum beginning with 1, with common ratio (1+i(12)/12):
$latex displaystyle sum_{k=0}^{11} (1+frac{i^{(12)}}{12})^{k}=frac{(1+frac{i^{(12)}}{12})^{12}-1}{frac{i^{(12)}}{12}}$
Multiplying by 45, and substituting the value we found for i(12)/12 gives us the final result:
$latex displaystyle 45sum_{k=0}^{11} (1+frac{i^{(12)}}{12})^{k}=45left(frac{(1+frac{i^{(12)}}{12})^{12}-1}{frac{i^{(12)}}{12}}right)approx 549.8298948$
Note: I used a few new commands here, such as dotsm for center ellipsis, sum for summation, and p style=”text-align:center” since the new version of html will get rid of the old align=”center” tag.
