The sum of the series frac{1}{{1 + {1^2} + {1 | vedclass

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(B) Let $T_n$ be the $n^{th}$ term of the series.$T_n = \frac{n}{{1 + n^2 + n^4}} = \frac{n}{{(1 + n^2)^2 - n^2}}$Using the identity $a^2 - b^2 = (a -

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$\frac{{n(n + 1)}}{{2({n^2} + n + 1)}}$

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(B) Let $T_n$ be the $n^{th}$ term of the series.$T_n = \frac{n}{{1 + n^2 + n^4}} = \frac{n}{{(1 + n^2)^2 - n^2}}$Using the identity $a^2 - b^2 = (a - b)(a + b)$:$T_n = \frac{n}{{(n^2 + n + 1)(n^2 - n + 1)}}$Using partial fractions:$T_n = \frac{1}{2} \left[ \frac{1}{{n^2 - n + 1}} - \frac{1}{{n^2 + n + 1}} \right]$Note that $n^2 - n + 1 = 1 + n(n - 1)$ and $n^2 + n + 1 = 1 + n(n + 1)$.Thus,$T_n = \frac{1}{2} \left[ \frac{1}{{1 + (n - 1)n}} - \frac{1}{{1 + n(n + 1)}} \right]$.Summing from $r=1$ to $n$:$S_n = \sum_{r=1}^n T_r = \frac{1}{2} \left[ \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{7} \right) + \dots + \left( \frac{1}{{1 + (n-1)n}} - \frac{1}{{1 + n(n+1)}} \right) \right]$This is a telescoping series,so all intermediate terms cancel out:$S_n = \frac{1}{2} \left[ 1 - \frac{1}{{1 + n(n + 1)}} \right] = \frac{1}{2} \left[ \frac{1 + n^2 + n - 1}{{n^2 + n + 1}} \right] = \frac{n(n + 1)}{{2(n^2 + n + 1)}}$.

The sum of the series $\frac{1}{{1 + {1^2} + {1^4}}} + \frac{2}{{1 + {2^2} + {2^4}}} + \frac{3}{{1 + {3^2} + {3^4}}} + \dots$ to $n$ terms is

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