Calculus: Early Transcendentals (2nd Edition)

$\dfrac {1\times \left( \dfrac {4}{25}\right) ^{21}-1}{\dfrac {4}{25}-1}=$
$\sum ^{20}_{k=0}\left( \dfrac {2}{5}\right) ^{2k}=a_{1}\dfrac {r^{n}-1}{r-1};a_{1}=1;r=\dfrac {4}{25};n=21\Rightarrow S_{n}=\dfrac {1\times \left( \dfrac {4}{25}\right) ^{21}-1}{\dfrac {4}{25}-1}=$ Using calculator as indicated in question we get