Question

You have $1000 to invest in an account and need to have $2000 in one year. What interest rate would you need to have in order to have this if the amount is compounded weekly? Round your answer to the nearest percent.

Answer 1

Answer:

$\large \boxed{\sf \ \ 70\% \ \ }$

Step-by-step explanation:

Hello,

We assume that the year is 52 weeks, and we note r the interest rate we are looking for. The rate is expressed in percent and is annually, meaning that the investment is, after the first week :

   $1000\cdot (1+\dfrac{r\%}{52})=1000\cdot (1+\dfrac{r}{5200})$

For the second week

   $1000\cdot (1+\dfrac{r}{5200})^2$

After 52 weeks

   $1000\cdot (1+\dfrac{r}{5200})^{52}$

and we want to be equal to 2000 so we need to solve:

$1000\cdot (1+\dfrac{r}{5200})^{52}=2000\\\\\text{*** divide by 1000 both sides ***}\\\\(1+\dfrac{r}{5200})^{52}=\dfrac{2000}{1000}=2\\\\\text{*** take the ln **}\\\\52\cdot ln(1+\dfrac{r}{5200})=ln(2)\\\\\text{*** divide by 52 ***}\\\\ln(1+\dfrac{r}{5200})=\dfrac{ln(2)}{52}\\\\\text{*** take the exp ***}\\\\\displaystyle 1+\dfrac{r}{5200}=exp(\dfrac{ln(2)}{52})=2^{(\dfrac{1}{52})}=\sqrt[52]{2}\\\\r = 5200\cdot (\sqrt[52]{2}-1)=69.77875...$

Rounded to the nearest percent, the solution is 70%.

If you want to double your capital in one year with weekly compounding you need an interest rate of 70% !!

Hope this helps.

Do not hesitate if you need further explanation.

Thank you