For kN, if the sum of the series 1+4k+8k^2+13k^3+19k^4+ is 10, then the value of k is ____…

Question

For $k\in\mathbb{N}$, if the sum of the series $1+\dfrac4k+\dfrac{8}{k^2}+\dfrac{13}{k^3}+\dfrac{19}{k^4}+\cdots$ is $10$, then the value of $k$ is \_\_\_\_\_\_\_\_.

Answer: 2

Accepted Answer

Using the successive-subtraction technique for arithmetico-geometric-type series (subtracting $	frac1k	imes$ the series repeatedly, since the numerators' second differences are constant), one obtains the equation
$9(k-1)^3 = 4k(k-1)+1,$
which is satisfied by $k=2$.

Solutions · 1