If ∑ _ k = 1 ^ 10 k k ^ 4 + k ^ 2 + 1 = (m)/(n) , where m and n are...

Given,

$\mathop{∑}_{k = 1}^{10}\dfrac{k}{{k}^{4} + {k}^{2} + 1} = \dfrac{m}{n}$

$⇒ \dfrac{1}{2}\mathop{∑}_{k = 1}^{10}\dfrac{\left({k}^{2} + k + 1\right) - \left({k}^{2} - k + 1\right)}{\left({k}^{2} + k + 1\right)\left({k}^{2} - k + 1\right)} = \dfrac{m}{n}$

\left\{\text{by using$the$formula}{x}^{4} + {x}^{2} + 1 = \left({x}^{2} + x + 1\right)\left({x}^{2} - x + 1\right)\right\}$</p><p>$⇒ \dfrac{1}{2}\left(\mathop{∑}_{k = 1}^{10}\left(\dfrac{1}{\left({k}^{2} - k + 1\right)} - \dfrac{1}{{k}^{2} + k + 1}\right)\right) = \dfrac{m}{n}$</p><p>$⇒ \dfrac{1}{2}\left(1 - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{7} + \dfrac{1}{7} - . . . . . . . . . - \dfrac{1}{111}\right) = \dfrac{m}{n}$</p><p>$⇒ \dfrac{1}{2}\left(1 - \dfrac{1}{111}\right) = \dfrac{m}{n}$</p><p>$⇒ \dfrac{55}{111} = \dfrac{m}{n}$</p><p>So,$m + n = 166