For k ∈ ℕ , if the sum of the series 1 + (4)/(k) + 8 k ^ 2 + 13 k ^...

Given,

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

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

On subtracting above two equations we get,

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

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

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

Again subtracting both equations we get,

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

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

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

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

$⇒ 9{k}^{3} - 31{k}^{2} + 31k - 10 = 0$

$⇒ k = 2$

Hence, $k = 2$