Is there any error in my solution : If $\sum^n_{r=1} r^4=I(n), $ then $\sum^n_{r=1}(2r-1)^4$ is equal to ..

Question

Problem :

If $\sum^n_{r=1} r^4=I(n), $ then $\sum^n_{r=1}(2r-1)^4$ is equal to

(a) $I(2n)-16I(n)$

(b) $I(3n)-2I(n)$

(c) $I(2n)-I(n)$

(d) $I(2n)+I(n)$

Please suggest if there is some error in the following solution of mine will be of great help.

Sol : $\sum^n_{r=1} r^4=I(n), $

$ I(2n) = 1^4+2^4+3^4 + \cdots +(2n-1)^4 +(2n)^4$

$ = 1^4 +3^4 +\cdots (2n-1)^4 + 2^4 (1^4 +2^4 + \cdots n^4)$

$ I(2n) = \sum^n_{r=1} (2r-1)^4 +16 I(n),$