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

Question

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

(a) f(2n) – 16f(n),∀ n ∈ N

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

(c) f(n)–16 \(f(\frac n2)\) n when n is odd

(d) None of these

Answer 1

Correct option is (a) f(2n) – 16f(n),∀ n ∈ N