Question

117. Sami opens an account and deposits $100 into it at the end of each month. The account earns 2% per year compounded monthly. Let Sn denote the amount of money in her account at the end of n months (just after she makes a deposit). For example, S1 = 100 and S2 = 100(1 + 0.02 /12)+100
c. Find an explicit function for Sn, and use it to find S12.

Answer 1

Answer:

The explicit function is:

$S_{n} =Pz^{n-1} +C(\frac{z^{n-1}-1}{z-1} )$

where

$z=(1+\frac{0.02}{12})$

and we calculate S12:

$S_{12} = 1211.06$

Step-by-step explanation:

Expanding a few steps of the compound interest:

$S_{1}=100\\S_{2}=S_{1}(1+\frac{0.02}{12})+100\\S_{3}=S_{2}(1+\frac{0.02}{12})+100\\...\\S_{n}=S_{n-1}(1+\frac{0.02}{12})+100$

We can write:

$(1+\frac{0.02}{12})=z$

$100=C$ for deposits

Then, expanding the previous equations would yield:

$S_{1}=C\\S_{2}=Cz + C\\S_{3}=Cz^{2}+Cz + C\\S_{n}=Cz^{n-1} +...+Cz+C$

This is a geometric series form, which can be simplified to:

$S_{n}=C(\frac{z^{n}-1}{z-1} )$

plugin in the values for the 12 month gives:

$S_{12}=100(\frac{(1+\frac{0.02}{12})^{12} -1}{\frac{0.02}{12}})=1211.0613$