Let ∑r=1n r4=f(n), then ∑r=1n (2r−1)4 is equal to

Let $\sum_{r = 1}^{n} r^{4} = f (n),$ then $\sum_{r = 1}^{n} (2 r - 1)^{4}$ is equal to

A
$f (2 n) - 16 f (n)$
B
$f (2 n) - 7 f (n)$
C
$f (2 n - 1) - 8 f (n)$
D
none of these

Solution:

We have

$\sum_{r = 1}^{n} (2 r - 1)^{4} = \sum_{r = 1}^{2 n} r^{4} - \sum_{r = 1}^{n} (2 r)^{4} = f (2 n) - 16 f (n)$

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