Binomial Theorem
Sum of Binomial Coefficients
Grade 11
Question:
<p>The value of \(2 \times {}^nC_1 + 2^3 \times {}^nC_3 + 2^5 \times {}^nC_5 + \ldots\) is</p>
<p>\(\dfrac{3^n + (-1)^n}{2}\)</p>
<p>\(\dfrac{3^n - (-1)^n}{2}\)</p>
<p>\(\dfrac{3^n + 1}{2}\)</p>
<p>\(\dfrac{3^n - 1}{2}\)</p>
Step-by-Step Solution
Key Concept: Use the binomial theorem with specific values of x to create equations that isolate odd-powered terms. By substituting x=2 and x=-2 into (1+x)^n, we can extract the sum of odd-indexed binomial coefficients multiplied by powers of 2.
<p><strong>Step 1: Write the binomial expansions</strong></p><p>We know that $(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \binom{n}{3}x^3 + \ldots$</p><p><strong>Step 2: Substitute x = 2</strong></p><p>$(1+2)^n = 3^n = \binom{n}{0} + \binom{n}{1}(2) + \binom{n}{2}(4) + \binom{n}{3}(8) + \binom{n}{4}(16) + \ldots$ ... (i)</p><p><strong>Step 3: Substitute x = -2</strong></p><p>$(1-2)^n = (-1)^n = \binom{n}{0} - \binom{n}{1}(2) + \binom{n}{2}(4) - \binom{n}{3}(8) + \binom{n}{4}(16) - \ldots$ ... (ii)</p><p><strong>Step 4: Subtract equation (ii) from equation (i)</strong></p><p>$3^n - (-1)^n = 2\binom{n}{1}(2) + 2\binom{n}{3}(8) + 2\binom{n}{5}(32) + \ldots$</p><p>$3^n - (-1)^n = 2 \times 2 \cdot \binom{n}{1} + 2 \times 2^3 \cdot \binom{n}{3} + 2 \times 2^5 \cdot \binom{n}{5} + \ldots$</p><p><strong>Step 5: Divide by 2</strong></p><p>$\dfrac{3^n - (-1)^n}{2} = 2\binom{n}{1} + 2^3\binom{n}{3} + 2^5\binom{n}{5} + \ldots$</p><p>This matches the given expression.</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B