Binomial Theorem
Binomial Series Summation
Grade 11
Question:
<p>\(\displaystyle\sum_{r=0}^{n} {}^nC_r \sin rx \cos(n-r)x\) is equal to</p>
<p>\(2^{n-1}\sin(n-1)x\)</p>
<p>\(2^n \sin nx\)</p>
<p>\(2^{n-1}\sin nx\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Recognize this sum as the imaginary part of the binomial expansion of (cos x + i sin x)^n = e^(inx), using the identity that binomial sum with trigonometric terms can be extracted via De Moivre's theorem.
<p><strong>Step 1:</strong> Use De Moivre's theorem. Note that (cos x + i sin x)^n = cos(nx) + i sin(nx)</p><p><strong>Step 2:</strong> Expand left side using binomial theorem: Σ(r=0 to n) ⁿCᵣ (cos x)^(n-r) (i sin x)^r</p><p><strong>Step 3:</strong> Simplify: Σ(r=0 to n) ⁿCᵣ i^r (cos x)^(n-r) (sin x)^r</p><p><strong>Step 4:</strong> Separate real and imaginary parts. The imaginary part gives us terms with i^r where r is odd.</p><p><strong>Step 5:</strong> For the given sum Σ ⁿCᵣ sin(rx) cos(n-r)x, use the product-to-sum conversion: sin(rx)cos(n-r)x = ½[sin(nx) + sin(2r-n)x]</p><p><strong>Step 6:</strong> Recognize the pattern. The sum equals <strong>2^(n-1) sin(nx)</strong> (or equivalently <strong>sin(nx)/2^(1-n)</strong> depending on form)</p><p>∴ Answer: <strong>C</strong></p>
Correct Answer: C