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Q.
Sum to $n$ terms of the series $\frac{1^{4}}{1.3}+\frac{2^{4}}{3.5}+\frac{3^{4}}{5.7}+.....$ is equal to
NTA AbhyasNTA Abhyas 2020Sequences and Series
Solution:
The $k^{t h}$ term of the given series is
$u_{k}=\frac{k^{4}}{\left(2 k - 1\right) \left(2 k + 1\right)}$ $=\frac{k^{2}}{4}+\frac{1}{16}+\frac{1}{32}\left(\frac{1}{2 k - 1} - \frac{1}{2 k + 1}\right)$
Therefore the sum of $n$ terms is
$S_{n}=u_{1}+u_{2}+....+u_{n}$
$=\frac{1}{4}\displaystyle \sum _{k = 1}^{n} k^{2} + \frac{n}{16}+\frac{1}{32}\left(1 - \frac{1}{2 n + 1}\right)$
$=\frac{n \left(n + 1\right) \left(2 n + 1\right)}{4.6}+\frac{n}{16}+\frac{n}{16 \left(2 n + 1\right)}$
$=\frac{n \left(n + 1\right) \left(n^{2} + n + 1\right)}{6 \left(2 n + 1\right)}$