Explanation

(b)
Given series is
$\frac{1}{1 + 1^{2} + 1^{4}} + \frac{2}{1 + 2^{2} + 2^{4}} + \frac{3}{1 + 3^{2} + 3^{4}} + ... + n$ terms
Let $T_{n}$ be the $n$th term of the series
$\frac{1}{1 + 1^{2} + 1^{4}} + \frac{2}{1 + 2^{2} + 2^{4}} + \frac{3}{1 + 3^{2} + 3^{4}} + ...$
Then, $T_{n} = \frac{n}{1 + n^{2} + n^{4}} = \frac{n}{\left( 1 + n^{2} \right)^{2} - n^{2}}$
$= \frac{n}{\left( n^{2} + n + 1 \right)(n^{2} - n + 1)}$
$= \frac{1}{2}\left\lbrack \frac{1}{n^{2} - n + 1} - \frac{1}{n^{2} + n + 1} \right\rbrack$
$= \frac{1}{2}\left\lbrack \frac{1}{1 + (n - 1)n} - \frac{1}{1 + (n + 1)} \right\rbrack$
$\therefore T_{1} = \frac{1}{2}\left\lbrack \frac{1}{1} - \frac{1}{1 + 1 \bullet 2} \right\rbrack$
$T_{2} = \frac{1}{2}\left\lbrack \frac{1}{1 + 1 \bullet 2} - \frac{1}{1 + 2 \bullet 3} \right\rbrack$
$T_{3} = \frac{1}{2}\left\lbrack \frac{1}{1 + 2 \bullet 3} - \frac{1}{1 + 3 \bullet 4} \right\rbrack$
$\ \cdots\ \cdots\ \cdots$
$\cdots\ \cdots\ \cdots$
$T_{n} = \frac{1}{2}\left\lbrack \frac{1}{1 + (n - 1)n} - \frac{1}{1 + (n + 1)} \right\rbrack$
Adding all these equations, we get
$\sum_{r = 1}^{n}{T_{r} = \frac{1}{2}\left\lbrack 1 - \frac{1}{1 + (n + 1)} \right\rbrack = \frac{n(n + 1)}{2(n^{2} + n + 1)}}$