the following is the problem I am trying to work on.
Kathryn deposits 100 into an account at the beggining of each 4 year period for 40 years. The account credits interest at an effective annual interest rate of $i$. The accumulated amount in the account at the end of 40 years is $X$. This is 5 times the accumulated amount in the account at the ned of 20 years. Calculate $X$.
The following is what I tried.
Let $1+i^*=(1+i)^4$ so that the calculation using annuity is a bit more straight forward.
Since Kathryn deposits at the beginning of the period this is an annuity-due with 1 extra year of conversion, so letting $Y$ be the accumulated amount at the end of 20 years we can find
$$Y=100\ddot s_{\overline {5}\rceil i^*}(1+i)$$
Similarlty I think that $X$ is
$$X=100\ddot s_{\overline {10}\rceil i^*}(1+i)$$
Solving for the equation $$5Y=X$$
I think I get
$$5 = \frac{(1+i^*)^{10}-1}{(1+i^*)^5-1}$$
Fortunately this is a difference of squares so we can simply solve for $i^*$ as
$$1+i^*= \sqrt[5]{4}$$
Up to here I am somewhat confident because the problem is from EXAM FM and this artificial looking situation seems how the problem was designed, but when I use this value of $i^*$
I end up getting
$$X = 100 \frac{15}{\sqrt[5]{4}-1} (\sqrt[4]{1+\sqrt[5]{4}})\approx 5793$$
but the answer is supposedly $6195$.
Maybe I fell for a common trap, but I cannot see what I did wrong. I appreciate your help.
Best Answer
Nice work. At the end you forgot, that $1+i^*=\sqrt[5]{4}=q^*$. It is not $i^*$. Thus there is no need to add 1 under the root sign. And additonal there is no need to take the fourth root.
The equation is $X=r\cdot \frac{\left( q^* \right)^{10}-1}{q^*-1}\cdot q^* =100\cdot \frac{\left( \sqrt[5]{4} \right)^{10}-1}{\sqrt[5]{4}-1}\cdot \sqrt[5]{4}\approx 6,195$
As you see, the formula looks very familiar-to both of us.