<p>If \(\sum_{r=1}^{n} r^4 = I(n)\), then \(\sum_{r=1}^{n} (2r-1)^4\) is equal to</p>
Step-by-Step Solution
Key Concept: Decompose the sum of fourth powers of odd numbers using the identity: sum of (2r-1)^4 = sum of all r^4 minus sum of even numbers' fourth powers, where even terms form a pattern with (2k)^4 = 16k^4.
<p><strong>Step 1:</strong> Express the required sum using the given I(n).</p><p>We need: Σ(2r-1)^4 for r=1 to n (summing fourth powers of first n odd numbers)</p><p><strong>Step 2:</strong> Use complementary series approach. Note that:</p><p>I(2n) = Σ(r^4) from r=1 to 2n = [sum of odd fourth powers] + [sum of even fourth powers]</p><p><strong>Step 3:</strong> For even terms, when r = 2k:</p><p>Σ(2k)^4 from k=1 to n = 16·Σ(k^4) from k=1 to n = 16·I(n)</p><p><strong>Step 4:</strong> Therefore:</p><p>Σ(2r-1)^4 from r=1 to n = I(2n) - 16·I(n)</p><p>∴ Answer: <strong>I(2n) - 16I(n)</strong></p>
Correct Answer: B