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March 29, 2024 at 12:10 pm #24346adminKeymaster
The sum of the series $ \frac{2}{3} + \frac{2}{9^2} + \frac{10}{9^3} + \frac{18}{9^4} + \frac{26}{9^5} + \cdots \infty $ is sought. To find this sum, we observe that the series can be represented in a more general form, which allows us to use the formula for the sum of an infinite geometric series.
First, let’s rewrite the given series:
\[
S = – \frac{2}{3} + \frac{2}{9^2} + \frac{10}{9^3} + \frac{18}{9^4} + \frac{26}{9^5} + \cdots
\]Noticing the pattern in the numerator, we can express the general term of the series as:
\[
a_n = \frac{8(n1) – 6}{9^n}, \quad \text{for } n \geq 2
\]Thus, the series becomes:
\[
S = – \frac{2}{3} + \sum_{n=2}^{\infty} \frac{8(n1) – 6}{9^n}
\]To simplify, we separate the series into two parts: one involving \(8(n1)/9^n\) and the other involving \(6/9^n\). This yields:
\[
S = – \frac{2}{3} + \left( \sum_{n=2}^{\infty} \frac{8(n1)}{9^n} \right) – \left( \sum_{n=2}^{\infty} \frac{6}{9^n} \right)
\]The first summation is a geometric series with the first term \(\frac{8}{9^2}\) and common ratio \(\frac{1}{9}\), and the second summation is also a geometric series but with the first term \(\frac{6}{9^2}\) and the same common ratio \(\frac{1}{9}\).
Using the formula for the sum of an infinite geometric series, \(S = \frac{a}{1 – r}\), where \(a\) is the first term and \(r\) is the common ratio, we calculate the sums of the two series:
\[
\sum_{n=2}^{\infty} \frac{8(n1)}{9^n} = \frac{\frac{8}{9^2}}{1 – \frac{1}{9}} = \frac{8}{72}
\]\[
\sum_{n=2}^{\infty} \frac{6}{9^n} = \frac{\frac{6}{9^2}}{1 – \frac{1}{9}} = \frac{6}{72}
\]Thus, the sum \(S\) becomes:
\[
S = – \frac{2}{3} + \frac{8}{72} – \frac{6}{72}
\]Simplifying further gives:
\[
S = – \frac{2}{3} + \frac{1}{9} – \frac{1}{12}
\]Finally, bringing the series to a common denominator and simplifying:
\[
S = \frac{24 + 4 – 3}{36} = \frac{23}{36}
\]Therefore, the sum of the series is \(\frac{23}{36}\).
 This topic was modified 4 months, 1 week ago by admin.
 This topic was modified 4 months, 1 week ago by admin.
 This topic was modified 4 months, 1 week ago by admin.
 This topic was modified 4 months, 1 week ago by admin.
 This topic was modified 4 months, 1 week ago by admin.
 This topic was modified 4 months, 1 week ago by admin.

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