Since each day, the amount of candy that Augustus will give away is 60% of the actual amount of candy, the amount of candy that has on day n depends on the amount of candy remaining from the day n-1. Let C_n be the amount of candy remaining on day n.
Since 60% of the candy from the day n-1 will be given away, then only 40% from the candy of the day n-1 will remain on day n. Then:
![\begin{gathered} C_n=(40)/(100)* C_(n-1) \\ =0.4C_(n-1) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/xibj46jfnd0u5kladkomldu4iohzwh1aby.png)
Since the amount of candy on day 1 is 100,000, then the recursive formula is:
![\begin{gathered} C_n=0.4C_(n-1) \\ C_1=100,000 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hbamv7ri9q794op1q7lk0na2uw5qaypuox.png)
After n-1 days, the initial amount of candies gets multiplied by a factor of 0.4 n-1 times. Then, the explicit formula for the amount of candies that remain on day n (after n-1 days) is:
![\begin{gathered} C_n=(0.4)^(n-1)* C_1 \\ =0.4^(n-1)*100,000 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2r3mdsigwzj99hyi52x6i4gjdji3qmx1s3.png)
Therefore, the answers are:
Explicit formula:
![C_n=0.4^(n-1)*100,000](https://img.qammunity.org/2023/formulas/mathematics/college/cb2h4cuyriutoialq4t9a4n7w145a4r5ob.png)
Recursive formula:
![\begin{gathered} C_n=0.4* C^{}_(n-1) \\ C_1=100,000 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wofzw1t8jb4rz56ldrq48ju5jkqe2oo8rd.png)