Annual simple interest vs. annual compound interest

The amount in a saving account after $n$ simple interest is given by 
\[P_n=P_0(1+nI),\]
and after $n$ annual compound interest is given by the formula 
\[P_n=P_0(1+I)^{n},\]
where $P_0$ is the initial balance and $I$ is the yearly interest rate. 

(a) If it had been gathering annual simple interest. Then 
\[P_{10}=P_0(1+10 I)=100(1+10I)=300  \Rightarrow 1+10I =3\]
\[\Rightarrow I=\frac{1}{5}= 20\%.\]

Now we compute the amount which have been in the account after only 5 years.

\[P_{5}=P_0(1+5 I)=100(1+5 \times \frac{1}{5})=200  \text{Euros}\]

(a) If it had been gathering annual interest, then $k=1$. So 
\[P_{10}=P_0(1+I)^{10}=100(1+I)^{10}=300\]
\[\Rightarrow (1+I)^{10}=3   \Rightarrow (1+I)=3^{\frac{1}{10}}.\]

Now we compute the ammount which have been in the account after only 5 years. 
\[P_{5}=P_0(1+I)^{5}=100(1+I)^{5}=100(3^{\frac{1}{10}})^{5}=100 \sqrt{3}=173  \text{Euros}.\]

(b) If it had been gathering annual interest, then $k=1$. So 
\[P_{10}=P_0(1+I)^{10}=100(1+I)^{10}=300\]
\[\Rightarrow (1+I)^{10}=3   \Rightarrow (1+I)=3^{\frac{1}{10}}.\]

Now we compute the amount which have been in the account after only 5 years. 
\[P_{5}=P_0(1+I)^{5}=100(1+I)^{5}=100(3^{\frac{1}{10}})^{5}=100 \sqrt{3}=173  \text{Euros}.\]

c) After 15 years, the savings account have more money in it if it is gathering annual compound interest. This is because after 10 years, there is 300 Euros in the account and during years 11-15 the account gathering annual compound interest will be gathering interest on the interests earned during years 11-15.