"For $k \in \mathbb{N}$, if the sum of the series $1+\frac{4}{k}+\frac{8}{k^{2}}+\frac{13}{k^{3}}+\frac{19}{k^{4}}+\ldots$ is 10 , then the value of $k$ is _________."

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"From the given series :

$10 = 1+\frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\ldots$

We isolate the 1 to get :

$9 = \frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\ldots$ .......(1)

Divide each term in the equation by $k$ :

$\frac{9}{k} = \frac{4}{k^2}+\frac{8}{k^3}+\frac{13}{k^4}+\ldots$ .......(2)

Subtracting the second equation from the first, we obtain :

$$9\left(1-\frac{1}{k}\right) = \frac{4}{k} + \frac{4}{k^2} + \frac{5}{k^3} + \frac{6}{k^4} + \ldots$$

This is equivalent to :

$$S = 9\left(1-\frac{1}{k}\right) = \frac{4}{k} + \frac{4}{k^2} + \frac{5}{k^3} + \frac{6}{k^4} + \ldots$$ .........(3)

Divide both sides by $k$ again, we get : $\frac{S}{k} = \frac{4}{k^2} + \frac{4}{k^3} + \frac{5}{k^4} + \ldots$ ..........(4)

Subtracting equation (4) from (3), we have :

$(1-\frac{1}{k})S = \frac{4}{k} + \frac{1}{k^3} + \frac{1}{k^4} + \ldots$

In this equation, you've treated the infinite series on the right-hand side as a geometric series with a ratio of $1/k$, so the sum of this series can be expressed as $\frac{1/k^3}{1 - 1/k}$.

Therefore, we get :

$9\left(1-\frac{1}{k}\right)^2 = \frac{4}{k} + \frac{1/k^3}{1 - 1/k}.$

Now, simplifying this equation, we get :

$9(k-1)^2 = 4k^2 - 4k + 1$

$\Rightarrow$$9(k^2 - 2k + 1) = 4k^2 - 4k + 1$

$\Rightarrow$$9k^2 - 18k + 9 = 4k^2 - 4k + 1$

$\Rightarrow$$9k^2 - 4k^2 - 18k + 4k + 9 - 1 = 0$

$\Rightarrow$$5k^2 - 14k + 8 = 0$

This is a standard form quadratic equation, $ax^2 + bx + c = 0$, where $a=5$, $b=-14$, and $c=8$.

The solutions can be found using the quadratic formula, $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Substituting the values for $a$, $b$, and $c$, we have :

$\Rightarrow$$k = \frac{14 \pm \sqrt{(-14)^2 - 4\times5\times8}}{2\times5}$

$\Rightarrow$$k = \frac{14 \pm \sqrt{196 - 160}}{10}$

$\Rightarrow$$k = \frac{14 \pm \sqrt{36}}{10}$

$\Rightarrow$$k = \frac{14 \pm 6}{10}$

So, the solutions are $k = 2$ or $k = 0.8$. However, since the question mentions $k \in \mathbb{N}$, i.e., $k$ is a natural number, the only valid solution is $k = 2$."

For $k \in \mathbb{N}$, if the sum of the series $1+\frac{4}{k}+\frac{8}{k^{2}}+\frac{13}{k^{3}}+\frac{19}{k^{4}}+\ldots$ is 10 , then the value of $k$ is _________.

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For $k \in \mathbb{N}$, if the sum of the series $1+\frac{4}{k}+\frac{8}{k^{2}}+\frac{13}{k^{3}}+\frac{19}{k^{4}}+\ldots$ is 10 , then the value of $k$ is _________.

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