Find an approximation of (0.99)5 using the first three terms of its expansion.

$0.99=1-0.01$

$\therefore(0.99)^{5}=(1-0.01)^{5}$

$={ }^{3} \mathrm{C}_{0}(1)^{5}-{ }^{3} \mathrm{C}_{1}(1)^{4}(0.01)+{ }^{5} \mathrm{C}_{2}(1)^{3}(0.01)^{2} \quad$ (Approximately)

$=1-5(0.01)+10(0.01)^{2}$

$=1-0.05+0.001$

$=1.001-0.05$

$=0.951$

Thus, the value of $(0.99)^{5}$ is approximately $0.951$.