$$ 1+3+3^{2}+...+3^{n-1}=\frac{3^{n}-1}{2} $$

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The sum of the geometric series $$ 1 + 3 + 3^2 + \ldots + 3^{n-1} $$ is $$ \frac{3^n - 1}{2} $$. To solve the problem, we need to prove the given formula for the sum of a geometric series: $$ 1 + 3 + 3^2 + \ldots + 3^{n-1} = \frac{3^n - 1}{2} $$ This is a geometric series with the first term $$ a = 1 $$ and the common ratio $$ r = 3 $$. The formula for the sum of the first $$ n $$ terms of a geometric series is: $$ S_n = \frac{a(1 - r^n)}{1 - r} $$ Let's apply this formula to our series: $$ S_n = \frac{1(1 - 3^n)}{1 - 3} $$ Now, simplify the denominator: $$ S_n = \frac{1 - 3^n}{-2} $$ To get the desired form, we can multiply both the numerator and the denominator by $$-1$$: $$ S_n = \frac{-(1 - 3^n)}{-2} $$ $$ S_n = \frac{3^n - 1}{2} $$ So, we have proven that the sum of the first $$ n $$ terms of the geometric series $$ 1 + 3 + 3^2 + \ldots + 3^{n-1} $$ is indeed equal to $$ \frac{3^n - 1}{2} $$.

\( 1+3+3^{2}+...+3^{n-1}=\frac{3^{n}-1}{2} \)

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\( 1+3+3^{2}+...+3^{n-1}=\frac{3^{n}-1}{2} \)

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