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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 :

guillermoreyesprince

Answer:

The explicit function is:

[tex]S_{n} =Pz^{n-1} +C(\frac{z^{n-1}-1}{z-1} )[/tex]

where

[tex]z=(1+\frac{0.02}{12}) [/tex]

and we calculate S12:

[tex]S_{12} = $1211.06[/tex]

Step-by-step explanation:

Expanding a few steps of the compound interest:

[tex]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[/tex]

We can write:

[tex](1+\frac{0.02}{12})=z[/tex]

[tex]100=C[/tex] for deposits

Then, expanding the previous equations would yield:

[tex]S_{1}=C\\S_{2}=Cz + C\\S_{3}=Cz^{2}+Cz + C\\S_{n}=Cz^{n-1} +...+Cz+C[/tex]

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

[tex]S_{n}=C(\frac{z^{n}-1}{z-1} )[/tex]

plugin in the values for the 12 month gives:

[tex]S_{12}=100(\frac{(1+\frac{0.02}{12})^{12} -1}{\frac{0.02}{12}})=1211.0613[/tex]

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