Let Σ r^4 = f(n); r=1, n , then Σ (2r - 1)^4 ; r=1, n is equal to

Question

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

(a) \(f(2n) - 16f(n), \forall n \in N\)

(b) \(f(n) - 16f(\frac{n-1}2)\) when n is odd

(c) \(f(n) - 16f(\frac{n}2)\) when n is odd

(d) None of these

Answer 1

Correct option is (a) \(f(2n) - 16f(n), \forall n \in N\)