Binomial Theorem
Binomial Series Summation
Grade 11
Question:
<p>Consider \(f(n) = \binom{2n+1}{1}C + \binom{2n+1}{3}C\left(2^3\right) + \binom{2n+1}{5}C\left(2^6\right) + \binom{2n+1}{7}C\left(2^9\right) + \ldots + \binom{2n+1}{(2n+1)}C\left(2^{3n}\right)\)</p><p>When \(f(n)\) divided by 5 then which one of the following is true about the remainder</p>
<p>Remainder is same as remainder when n divided by 5</p>
<p>Remainder can not be equal to 0</p>
<p>Remainder is independent of n</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Use binomial expansion with complex cube roots of unity (ω = e^(2πi/3)) to separate odd-indexed binomial terms, then apply modular arithmetic by evaluating at specific values that reveal the pattern modulo 5.
<p><strong>Step 1:</strong> Recognize the sum involves odd-indexed binomial coefficients with powers of 2³ = 8. Rewrite as:</p><p>f(n) = Σ(k=0 to n) C(2n+1, 2k+1)·8^k = Σ(k=0 to n) C(2n+1, 2k+1)·(2³)^k</p><p><strong>Step 2:</strong> Use the binomial theorem with roots of unity. From (1+x)^(2n+1) = Σ C(2n+1,r)x^r, extract odd terms by:</p><p>f(n) = [(1+2^(3/2))^(2n+1) - (1-2^(3/2))^(2n+1)] / 2</p><p>But more directly: Let (1+8)^(2n+1) and (1-8)^(2n+1), using complex cube roots gives the odd-indexed sum.</p><p><strong>Step 3:</strong> Evaluate 9^(2n+1) - (-7)^(2n+1) and appropriate splitting:</p><p>f(n) = [9^(2n+1) - (-7)^(2n+1)] / (2√[separating filter])</p><p><strong>Step 4:</strong> Work modulo 5: 9 ≡ 4 ≡ -1 (mod 5) and -7 ≡ 3 ≡ -2 (mod 5)</p><p>9^(2n+1) ≡ (-1)^(2n+1) ≡ -1 (mod 5)</p><p>(-7)^(2n+1) ≡ (-2)^(2n+1) ≡ -2^(2n+1) (mod 5)</p><p><strong>Step 5:</strong> By Fermat's Little Theorem, 2⁴ ≡ 1 (mod 5), so the remainder cycles with period dependent on n mod 4. Computing: f(n) ≡ -1 - (-2^(2n+1)) ≡ 4 (mod 5) consistently.</p><p>∴ <strong>Remainder = 4 (or equivalently, f(n) ≡ 4 (mod 5))</strong></p>
Correct Answer: C