Using the formula for the compound interest, we have:
[tex]\begin{gathered} A=P(1+i)^n\text{ (A: amount, P:principal, i:interest, n:years}) \\ A=2000(1+0.06)^4\text{ (Replacing the values)} \\ A=2000(1.06)^4\text{ (Adding)} \\ A=2000\text{ (1.26) (Raising 1.06 to the power of 4)} \\ A=2524.95\text{ (Multiplying)} \\ \text{The answer is \$2524.95} \end{gathered}[/tex]