Answer :
Answer:
Beatrice will accumulate $1230.72 at the end of the year.
Step-by-step explanation:
We can write:
[tex](1+\frac{0.055}{12})=z[/tex]
[tex]100=C[/tex] for deposits
The first month would have only the deposit reflected in her balance, then, expanding some steps of the calculation would yield:
[tex]S_{1}=C\\S_{2}=Cz + C\\S_{3}=Cz^{2}+Cz + C\\S_{n}=Cz^{n-1} +...+Cz+C[/tex]
A geometric series is given by:
[tex]1+x+...+x^{m}=\frac{x^{m+1}-1}{x-1}[/tex]
Translating our series to the short form:
[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{0.055}{12})^{12} -1}{\frac{0.055}{12}})=1230.7169[/tex]