Animated Solution for Mathematics - Sequence and Series: For $k\in \mathbb{N}$, if the sum of the series $1+\frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\dots$ is 10, then the value of $k$ is

The given series is $S=1+\frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\cdots =10$. This is a second-order Arithmetico-Geometric series because the differences of the numerators ($4-1=3$, $8-4=4$, $13-8=5$, $19-13=6$) form an Arithmetic Progression.

To solve this, we must systematically reduce the complexity of the series by peeling back its layers using the Shift-and-Subtract method.

We write the series $S$ and multiply it by the common ratio $\frac{1}{k}$ to obtain $\frac{S}{k}$. Subtracting $\frac{S}{k}$ from $S$ yields:

The new numerators ($1,3,4,5,6$) are simpler, but the arithmetic progression persists. We must apply the method once more to reach the core.

We apply the shift-and-subtract technique again to the resulting series. Multiplying by $\frac{1}{k}$ and subtracting once more transforms the expression into:

The term inside the brackets is a standard infinite Geometric Progression with first term $a=\frac{1}{k^2}$ and common ratio $r=\frac{1}{k}$.

Using the infinite GP sum formula $\frac{a}{1-r}$, the bracketed term simplifies to:

Substituting this back into our equation with $S=10$, we get:

Expanding and simplifying this expression leads to the cubic equation:

Testing for integer roots, we find that for $k=2$:

Thus, the value of $k$ is 2.