91Ó°ÊÓ

Given a geometric series, we can use the formula: $$ S_n =\frac{a_1 (1 - r^n)}{1 - r} $$ Where: - \(S_n\) is the sum of the series up to the nth term; - \(a_1\) is the first term in the series; - \(r\) is the common ratio; - and \(n\) is the number of terms in the series. Given the summation expression: $$ \sum_{k=1}^{40} \frac{3}{2^k} $$ We can recognize that this is a geometric series with the following properties: - First term (\(a_1\)): When k = 1, the first term is \(\frac{3}{2^1}\), which equals \(\frac{3}{2}\). So \(a_1 = \frac{3}{2}\). - Common ratio (\(r\)): The common ratio between consecutive terms is \(\frac{1}{2}\) (as each term is divided by 2). - Number of terms (\(n\)): Since we are summing from k=1 to k=40, there are 40 terms.

Now that we have the first term, common ratio, and number of terms, we can plug these values into our summation formula: $$ S_n =\frac{a_1 (1 - r^n)}{1 - r} $$ $$ S_{40} = \frac{\frac{3}{2}(1 - (\frac{1}{2})^{40})}{1 - \frac{1}{2}} $$

Now, we need to simplify the expression: $$ S_{40} = \frac{\frac{3}{2}(1 - (\frac{1}{2})^{40})}{\frac{1}{2}} $$ Multiply both numerator and denominator by 2: $$ S_{40} = \frac{3(1 - (\frac{1}{2})^{40})}{1} $$ $$ S_{40} = 3(1 - (\frac{1}{2})^{40}) $$ Therefore, the sum of the given geometric series is: $$ S_{40} = 3(1 - (\frac{1}{2})^{40}) $$