First slide

Sigma operation series

Question

If $\frac{1^{4}}{13} + \frac{2^{4}}{35} + \frac{3^{4}}{5 .7} + \dots + \frac{n^{4}}{(2 n - 1) (2 n + 1)} = \frac{1}{48} f (n) + \frac{n}{16 (2 n + 1)},$ then f (n) is equal to

Moderate

A

$n (4 n^{2} + 3 n + 2)$
B

$n (4 n^{2} + 6 n + 5)$
C

$n (4 n^{2} + 5 n + 6)$
D

None of these

Solution

We have, $\begin{array}{l} t_{n} = \frac{n^{4}}{(2 n - 1) (2 n + 1)} \\ = \frac{n^{2}}{4} + \frac{1}{16} + \frac{1}{16 (4 n^{2} - 1)} \\ = \frac{4 n^{2} + 1}{16} + \frac{1}{32} (\frac{1}{2 n - 1} - \frac{1}{2 n + 1}) \end{array}$
$\begin{array}{l} ∴ S_{n} = \sum_{n = 1}^{n} t_{n} \\ = \sum_{n = 1}^{n} \frac{4 n^{2} + 1}{16} + \frac{1}{32} \sum_{n = 1}^{n} (\frac{1}{2 n - 1} - \frac{1}{2 n + 1}) \\ = \frac{1}{4} \frac{n (n + 1) (2 n + 1)}{6} + \frac{1}{16} n + \frac{1}{32} (1 - \frac{1}{2 n + 1}) \\ = \frac{n}{48} (4 n^{2} + 6 n + 5) + \frac{1}{32} (1 - \frac{1}{2 n + 1}) \\ = \frac{n (4 n^{2} + 6 n + 5)}{48} + \frac{n}{16 (2 n + 1)} \\ ∴ f (n) = n (4 n^{2} + 6 n + 5) \end{array}$

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