Find the value of $$ \sum_{n=1}^{\infty}\left(\frac{9}{2^{n}}-\frac{7}{6^{n}}\right) $$ if $$ \sum_{n=1}^{\infty} \frac{1}{2^{n}}=1 $$ and $$ \sum_{n=1}^{\infty} \frac{1}{6^{n}}=\frac{1}{5} $$

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Given the series

$$
\sum_{n=1}^{\infty}\left(\frac{9}{2^{n}} - \frac{7}{6^{n}}\right),
$$

we can split it into two separate sums:

$$
\sum_{n=1}^{\infty}\frac{9}{2^{n}} - \sum_{n=1}^{\infty}\frac{7}{6^{n}}.
$$

Since constants can be factored out, we have

$$
9\sum_{n=1}^{\infty}\frac{1}{2^{n}} - 7\sum_{n=1}^{\infty}\frac{1}{6^{n}}.
$$

We are given:

$$
\sum_{n=1}^{\infty}\frac{1}{2^{n}} = 1 \quad \text{and} \quad \sum_{n=1}^{\infty}\frac{1}{6^{n}} = \frac{1}{5}.
$$

Substitute these values into the expression:

$$
9(1) - 7\left(\frac{1}{5}\right) = 9 - \frac{7}{5}.
$$

To combine the terms, write $$9$$ as $$\frac{45}{5}$$:

$$
\frac{45}{5} - \frac{7}{5} = \frac{38}{5}.
$$

Thus, the value of the series is

$$
\frac{38}{5}.
$$
The value of the series is $$ \frac{38}{5} $$.

Find the value of if and

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