Question

Evaluate (0.99)${}^{15}$ and (1.0025)${}^{10}$ correct to four decimal places.

Answer 1

$(.99{)}^{15}={(1-.01)}^{15}$
$=1{-}^{15}{C}_1(.01)+{}^{15}{C}_2{(.01)}^2-{}^{15}{C}_3{(.01)}^3+......$
We want the answer correct to only 4 decimal places and as such we have left further expansion.
$(.99{)}^{15}=(1-.01{)}^{15}$
$=1-15\left(0.1\right)+\frac{15.14}{1.2}\left(.0001\right)-\frac{\mathrm{15.14.13}}{\mathrm{1.2.3}}\left(.000001\right)$
$=1-.15+0105-.000455$
$=1.0105-\left(.150455\right)=1.0105-.1504=0.8601$