### Instruction: Let $k$ be a real number such that $k > 1$ and
\[\sum_{n=1}^{\infty} \frac{5n-1}{k^n} = \frac{13}{4}.\]Find $k.$.### Response: I notice that this is an infinite geometric series, where the ratio $r$ between consecutive terms is $\frac{1}{k}.$
I recall that an infinite geometric series converges to a finite sum if and only if $|r| < 1.$
Since $k > 1,$ this condition is satisfied.
I also recall that the formula for the sum of an infinite geometric series is $\frac{a}{1-r},$ where $a$ is the first term of the series.
Therefore, I can write
\[\frac{13}{4} = \frac{a}{1-r} = \frac{a}{1 - \frac{1}{k}} = \frac{ak}{k-1}.\]
To find $a,$ I need to plug in $n = 1$ into the general term of the series.
This gives
\[a = \frac{5(1) - 1}{k^1} = \frac{4}{k}.\]
Substituting this into the equation, I get
\[\frac{13}{4} = \frac{ak}{k-1} = \frac{4}{k} \cdot \frac{k}{k-1} = \frac{4}{k-1}.\]
Multiplying both sides by $k-1,$ I get
\[13k - 13 = 16.\]
Adding 13 to both sides, I get
\[13k = 29.\]
Dividing both sides by 13, I get
\[k = \frac{29}{13}.\]