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

Given,

$10=1+\frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}........\infty .....\left(1\right)$

$\frac{10}{k}=\frac{1}{k}+\frac{4}{k^2}+\frac{8}{k^3}+\frac{13}{k4}+........\infty ....\left(2\right)$

On subtracting above two equations we get,

$10-\frac{10}{k}=1+\frac{3}{k}+\frac{4}{k^2}+\frac{5}{k^3}+\frac{6}{k^4}........\infty$

$\Rightarrow 10\left(1-\frac{1}{k}\right)=1+\frac{3}{k}+\frac{4}{k^2}+\frac{5}{k^3}+\frac{6}{k^4}........\infty .....\left(3\right)$

$\Rightarrow 10\left(1-\frac{1}{k}\right)\left(\frac{1}{k}\right)= \frac{1}{k}+\frac{3}{k^2}+\frac{4}{k^3}+\frac{5}{k^4}+\frac{6}{k^5}........\infty .....\left(4\right)$

Again subtracting both equations we get,

$\Rightarrow 10{\left(1-\frac{1}{k}\right)}^2=1+\frac{2}{k}+\frac{1}{k^2}+\frac{1}{k^3}+\frac{1}{k^4}........\infty$

$\Rightarrow 10{\left(1-\frac{1}{k}\right)}^2=\left(1+\frac{1}{k}+\frac{1}{k^2}+\frac{1}{k^3}+\frac{1}{k^4}........\infty \right) +\frac{1}{k}$

$\Rightarrow 10{\left(1-\frac{1}{k}\right)}^2=\left(\frac{1}{1-{\displaystyle \frac{1}{k}}}\right) +\frac{1}{k}$

$\Rightarrow 10{\left(k-1\right)}^3=k\left(k^2+k-1\right)$

$\Rightarrow 9k^3-31k^2+31k-10=0$

$\Rightarrow k=2$ 

Hence, $k=2$