Find the sum of the first n terms and the sum | vedclass

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Question

(A) Given the geometric series $1 + \frac{2}{3} + \frac{4}{9} + \dots$ Here,the first term $a = 1$ and the common ratio $r = \frac{2/3}{1} = \frac{2}{

Vedclass · Accepted Answer

$S_n = 3[1 - (\frac{2}{3})^n], S_5 = \frac{211}{81}$

Vedclass · Answer

(A) Given the geometric series $1 + \frac{2}{3} + \frac{4}{9} + \dots$ Here,the first term $a = 1$ and the common ratio $r = \frac{2/3}{1} = \frac{2}{3}$. Since $|r| Substituting the values,$S_n = \frac{1(1 - (2/3)^n)}{1 - 2/3} = \frac{1 - (2/3)^n}{1/3} = 3[1 - (\frac{2}{3})^n]$. For the sum of the first $5$ terms,$n = 5$: $S_5 = 3[1 - (\frac{2}{3})^5] = 3[1 - \frac{32}{243}] = 3[\frac{243 - 32}{243}] = 3[\frac{211}{243}] = \frac{211}{81}$.

Find the sum of the first $n$ terms and the sum of the first $5$ terms of the geometric series $1 + \frac{2}{3} + \frac{4}{9} + \dots$

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