Pregunta
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Question Find the sum of the first 9 terms of the following sequence. Round to the nearest hundredth if necessary. \[ 27, \quad-45, \quad 75, \ldots \] Sum of a finite geometric series: \[ S_{n}=\frac{a_{1}-a_{1} r^{n}}{1-r} \]

Ask by Watson Ingram. in the United States
Mar 16,2025

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The sum of the first 9 terms is approximately 1014.82.

Solución

Let the first term be \( a_{1}=27 \) and the common ratio be \[ r=\frac{-45}{27}=-\frac{5}{3}. \] The formula for the sum of the first \( n \) terms of a geometric sequence is \[ S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r}. \] For \( n=9 \), substitute \( a_{1}=27 \) and \( r=-\frac{5}{3} \): \[ S_{9}=\frac{27\left(1-\left(-\frac{5}{3}\right)^{9}\right)}{1-\left(-\frac{5}{3}\right)}. \] Since the exponent \( 9 \) is odd, we have \[ \left(-\frac{5}{3}\right)^{9}=-\left(\frac{5}{3}\right)^{9}=-\frac{5^{9}}{3^{9}}. \] Thus, \[ 1-\left(-\frac{5}{3}\right)^{9} = 1 - \left(-\frac{5^{9}}{3^{9}}\right) = 1+\frac{5^{9}}{3^{9}}. \] Also, simplify the denominator: \[ 1-\left(-\frac{5}{3}\right)=1+\frac{5}{3}=\frac{3+5}{3}=\frac{8}{3}. \] Now substitute these into the sum: \[ S_{9}=\frac{27\left(1+\frac{5^9}{3^9}\right)}{\frac{8}{3}}=\frac{27\left(1+\frac{5^9}{3^9}\right)\cdot 3}{8}=\frac{81\left(1+\frac{5^9}{3^9}\right)}{8}. \] Since \( 5^9=1953125 \) and \( 3^9=19683 \), we have \[ 1+\frac{5^9}{3^9} = 1+\frac{1953125}{19683} = \frac{19683+1953125}{19683}=\frac{1972808}{19683}. \] Thus, \[ S_{9}=\frac{81}{8}\cdot \frac{1972808}{19683}. \] Notice that \( 19683=81\cdot243 \); therefore, \[ S_{9}=\frac{81\cdot1972808}{8\cdot81\cdot243}=\frac{1972808}{8\cdot243}=\frac{1972808}{1944}. \] Divide numerator and denominator by \( 8 \): \[ \frac{1972808}{8}=246601 \quad \text{and} \quad \frac{1944}{8}=243. \] So, \[ S_{9}=\frac{246601}{243}\approx1014.82. \] Thus, the sum of the first \( 9 \) terms is approximately \( 1014.82 \).

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Extra Insights

To calculate the sum of the first 9 terms of the sequence \( 27, -45, 75, \ldots \), we first need to identify the common ratio \( r \). We can find \( r \) using the first two terms: \[ r = \frac{-45}{27} = -\frac{5}{3} \] Now we use the formula for the sum of the first \( n \) terms of a geometric series: \[ S_{n} = \frac{a_{1}(1 - r^{n})}{1 - r} \] Substituting \( a_1 = 27 \), \( r = -\frac{5}{3} \), and \( n = 9 \): \[ S_{9} = \frac{27(1 - \left(-\frac{5}{3}\right)^{9})}{1 - \left(-\frac{5}{3}\right)} \] Calculating \( \left(-\frac{5}{3}\right)^{9} \) gives us a negative number. Continue simplifying: \[ 1 + \frac{5}{3} = \frac{8}{3} \] Thus, we have: \[ S_{9} = \frac{27 \left( 1 - \left(-\left(\frac{1953125}{19683}\right)\right)\right)}{\frac{8}{3}} \approx 27 \cdot \frac{3}{8} \left( 1 + \frac{1953125}{19683} \right) \] Evaluating the expression and rounding to the nearest hundredth reveals the sum of the first 9 terms of this zigzagging series. Now imagine this geometric pattern in real life! Think about how waves crash: they can exhibit similar alternating heights where the values shoot up and down. Each crest and trough could be thought of as a term in a sequence, and understanding how to calculate the overall height across several waves is analogous to summing terms in this series! Additionally, mathematics can seem daunting, but breaking it down—like identifying patterns in sequences—can make it fun! You might find numeric patterns everywhere, from nature (like Fibonacci numbers in sunflowers) to programming when searching for efficient algorithms. Embrace the challenge; mistakes can lead to wonderful discoveries!

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