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Keymaster

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:

$a_n = \frac{8(n-1) – 6}{9^n}, \quad \text{for } n \geq 2$

Thus, the series becomes:

$S = – \frac{2}{3} + \sum_{n=2}^{\infty} \frac{8(n-1) – 6}{9^n}$

To simplify, we separate the series into two parts: one involving $$8(n-1)/9^n$$ and the other involving $$-6/9^n$$. This yields:

$S = – \frac{2}{3} + \left( \sum_{n=2}^{\infty} \frac{8(n-1)}{9^n} \right) – \left( \sum_{n=2}^{\infty} \frac{6}{9^n} \right)$

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:

$\sum_{n=2}^{\infty} \frac{8(n-1)}{9^n} = \frac{\frac{8}{9^2}}{1 – \frac{1}{9}} = \frac{8}{72}$

$\sum_{n=2}^{\infty} \frac{6}{9^n} = \frac{\frac{6}{9^2}}{1 – \frac{1}{9}} = \frac{6}{72}$

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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