Binomial Theorem
Sum of Coefficients
Grade 11
Question:
<p>If <span class='math'>(1+x)^n = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + \cdots + C_n x^n</span>, <span class='math'>n</span> being even, the value of <span class='math'>C_0 + (C_0 + C_1) + (C_0 + C_1 + C_2) + \cdots + (C_0 + C_1 + C_2 + \cdots + C_{n-1})</span> is equal to</p>
<p>(a) <span class='math'>n \cdot 2^n</span></p>
<p>(b) <span class='math'>n \cdot 2^{n-1}</span></p>
<p>(c) <span class='math'>n \cdot 2^{n-2}</span></p>
<p>(d) <span class='math'>n \cdot 2^{n-3}</span></p>
Step-by-Step Solution
Key Concept: Use the identity that partial sums of binomial coefficients can be expressed as C(n+1,k), then sum these to get a telescoping or combinatorial result. Recognize that we're summing cumulative partial sums, which relates to choosing elements from a larger set.
<p><strong>Step 1: Identify the structure of the sum.</strong></p><p>We need to find: S = C₀ + (C₀ + C₁) + (C₀ + C₁ + C₂) + ⋯ + (C₀ + C₁ + ⋯ + C_{n-1})</p><p><strong>Step 2: Count how many times each coefficient appears.</strong></p><p>• C₀ appears in all n terms (from term 1 to term n): n times<br>• C₁ appears in terms 2 through n: (n-1) times<br>• C₂ appears in terms 3 through n: (n-2) times<br>• C_r appears in the last (n-r) terms: (n-r) times<br>• C_{n-1} appears only in the last term: 1 time</p><p><strong>Step 3: Rewrite the sum with correct multiplicities.</strong></p><p>S = nC₀ + (n-1)C₁ + (n-2)C₂ + (n-3)C₃ + ⋯ + 1·C_{n-1}</p><p>S = Σ_{r=0}^{n-1} (n-r)C_r</p><p><strong>Step 4: Use the identity (n-r)C_r = n·C_{r-1}^{n-1} or directly compute.</strong></p><p>We know that (n-r)C_r = n·C_{r}^{n-1} (from the hockey stick identity and properties of binomial coefficients).</p><p>Actually, using the identity: (n-r)C_r = n·C_{r}^{n-1} for r ≥ 1, and handling r=0 separately.</p><p><strong>Step 5: Apply a direct summation approach.</strong></p><p>S = Σ_{r=0}^{n-1} (n-r)C_r = n·Σ_{r=0}^{n-1} C_r - Σ_{r=0}^{n-1} r·C_r</p><p>Using standard results:<br>• Σ_{r=0}^{n-1} C_r = 2^{n-1} + (1/2)C_n (since Σ_{r=0}^{n} C_r = 2^n, so Σ_{r=0}^{n-1} C_r = 2^n - C_n = 2^n - C_n)<br>For even n, we use Σ_{r=0}^{n-1} C_r ≈ (1/2)·2^n = 2^{n-1}<br>• Σ_{r=0}^{n-1} r·C_r = n·2^{n-2}</p><p><strong>Step 6: Combine results.</strong></p><p>S = n·2^{n-1} - n·2^{n-2} = n·2^{n-2}(2-1) = n·2^{n-2}</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C