Diketahui:

Deret geometri $\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\dots$

Ditanya:

Jumlah empat suku pertama deret tersebut?

Jawab:

Deret geometri tersebut memiliki:

$a=\frac{1}{4}$

$r=\frac{U_{n+1}}{U_n}=\frac{U_2}{U_1}=\frac{\frac{1}{12}}{\frac{1}{4}}=\frac{1}{12}.\frac{4}{1}=\frac{1}{3}$

Jumlah 4 suku pertama deret geometri dengan $r<1$ adalah

$S_n=\frac{a\left(1-r^n\right)}{1-r}$

$S_4=\frac{\frac{1}{4}\left(1-\left(\frac{1}{3}\right)^4\right)}{1-\frac{1}{3}}$

$S_4=\frac{\frac{1}{4}\left(1-\frac{1}{3^4}\right)}{\frac{3-1}{3}}$

$S_4=\frac{\frac{1}{4}\left(1-\frac{1}{81}\right)}{\frac{2}{3}}$

$S_4=\frac{1}{4}\left(\frac{81}{81}-\frac{1}{81}\right).\frac{3}{2}$

$S_4=\frac{1}{4}.\frac{3}{2}\left(\frac{81-1}{81}\right)$

$S_4=\frac{3}{8}\left(\frac{80}{81}\right)$

$S_4=\frac{10}{27}$