The sum to 10 terms of the series 1 1 + 1 ^ 2 + 1 ^ 4 + 2 1 + 2 ^ 2...

We have,

$\dfrac{1}{1 + {1}^{2} + {1}^{4}} + \dfrac{2}{1 + {2}^{2} + {2}^{4}} + \dfrac{3}{1 + {3}^{2} + {3}^{4}} + \ldots$

Here,

${t}_{n} = \dfrac{n}{1 + {n}^{2} + {n}^{4}}$

$= \dfrac{n}{( {n}^{2} + n + 1 ) ( {n}^{2} - n + 1 )}$

So,

${t}_{1} = \dfrac{1}{2}\left(1 - \dfrac{1}{3}\right)$

${t}_{2} = \dfrac{1}{2}\left(\dfrac{1}{3} - \dfrac{1}{7}\right)$

$⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮$

${t}_{10} = \dfrac{1}{2}\left(\dfrac{1}{91} - \dfrac{1}{111}\right)$

Now,

${S}_{10} = {t}_{1} + {t}_{2} + {t}_{3} + . . . . . + {t}_{10}$

$⇒ {S}_{10} = \dfrac{1}{2}\left(1 - \dfrac{1}{111}\right)$

$⇒ {S}_{10} = \dfrac{55}{111}$