Compounded interest compounds

Interest earned is given by: $I = A(1+r)^n - A$ , where $A$ is the principal amount, $r$ is the interest rate, and $n$ is the number of periods.

Therefore, after $n=2$ years, the interest earned by:

$\begin{cases} \text{Alex: } & I_A & = 1000(1+0.10)^2 - 1000 & = 210 \\ \text{Brian: } & I_B & = 2000(1+0.05)^2 - 1000 & = 205 \\ \text{Difference: } & I_A-I_B & = 210 - 205 & = \boxed{5} \end{cases}$

For Alex : 1000 x (1.10^2) Where 1000 is Alex's amount and the 1.10 is the percentage multiplier, basically 1 + 0.10. ^2 is the amount of years, in this case it would be two After calculating 1000 x (1.10^2) this would give us 1210 we would then take away 1000, the initial amount, because we only want the interest and this would give us 210

For Brian : 2000 x (1.05^2) = 2205 for this we use the same terminology as we did for Alex. So now take away the initial amount which is 2000 and that would give us 205

For the difference or answer : We take away Alex's answer from Brian's which would be 210 - 205 = 5 and that's our answer.

Formula: $F=P(1+i)^n$

For Alex: $F=1000(1+0.1)^2=1210$

So the interest earned for Alex is $1210-1000=210$

For Brian: $F=2000(1+0.05)^2=2205$

So the interest earned for Brian is $2205-2000=205$

The difference between the interest earned is $210-205=\color{#D61F06}\boxed{5}$