Question:
Find an approximation of $(0.99)^{5}$ using the first three terms of its expansion.
Solution:
$0.99=1-0.01$
$\therefore(0.99)^{5}=(1-0.01)^{5}$
$={ }^{3} \mathrm{C}_{0}(1)^{5}-{ }^{3} \mathrm{C}_{1}(1)^{4}(0.01)+{ }^{5} \mathrm{C}_{2}(1)^{3}(0.01)^{2} \quad$ (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 $0.951$.