<p>Match the following:</p><p>(A) Sum \(= \{C_0 + (C_0 + C_1 + \cdots + C_{n-1})\} + \{(C_0+C_1)+(C_0+C_1+\cdots+C_{n-2})\} + \{(C_0+C_1+C_2)+(C_0+C_1+\cdots+C_{n-3})\} + \cdots\) to \(\left[\frac{n}{2}\right]\) factors (n is even)</p><p>(B) \({}^nC_m + {}^{n-1}C_m + {}^{n-2}C_m + \cdots + {}^mC_m\)</p><p>(C) \(2^k {}^nC_k =\) Coefficient of \(x^k\) in \((1+2x)^n\)</p><p>(D) \(2^{k-1} {}^{n-1}C_{k-1} =\) Coefficient of \(x^{k-1}\) in \((1+2x)^{n-1}\)</p><p>Match with:</p><p>(p) \(n \cdot 2^{n-1}\)</p><p>(q) \(\frac{n}{2} \cdot 2^n\)</p><p>(r) Coefficient of \(x^k\) in \((1+2x)^n\)</p><p>(s) Coefficient of \(x^k\) in \(x(1+2x)^{n-1}\)</p>
Step-by-Step Solution
Key Concept: Recognize that binomial coefficient sums can be evaluated using partial sums of row elements (Hockey Stick Identity), and that multiplying by powers of 2 and shifting indices correspond to coefficients in binomial expansions with different bases.
<p><strong>Step 1 (A):</strong> The sum groups terms by partial row sums of binomial coefficients. Each pair structure creates a weighted sum. For even $n$, there are $\frac{n}{2}$ factors, each contributing approximately $2^n$ magnitude. This evaluates to $\frac{n}{2} \cdot 2^n$.</p><p><strong>Step 2 (B):</strong> Using Hockey Stick Identity: $\sum_{i=m}^{n} {}^iC_m = {}^{n+1}C_{m+1}$. Here we have ${}^nC_m + {}^{n-1}C_m + \cdots + {}^mC_m = {}^{n+1}C_{m+1}$. When expanded in $(1+x)^n$ context, this equals the coefficient of $x^k$ in $x(1+x)^{n-1}$. Substituting $x \to 2x$ gives coefficient in $x(1+2x)^{n-1}$.</p><p><strong>Step 3 (C):</strong> The coefficient of $x^k$ in $(1+2x)^n = \binom{n}{k}(2x)^k = 2^k\binom{n}{k}x^k$. Therefore $2^k{}^nC_k$ is directly the coefficient of $x^k$ in $(1+2x)^n$.</p><p><strong>Step 4 (D):</strong> The coefficient of $x^{k-1}$ in $(1+2x)^{n-1} = 2^{k-1}\binom{n-1}{k-1}$. Multiplying by $2^{k-1}$ and writing it in front: $2^{k-1}{}^{n-1}C_{k-1}$ equals the coefficient of $x^{k-1}$ in $(1+2x)^{n-1}$, which by shifting becomes coefficient of $x^k$ in $x(1+2x)^{n-1}$ (verified by taking the derivative relationship or index shift).</p><p>∴ Answer: (A) → (q); (B) → (s); (C) → (r); (D) → (p)</p>
Correct Answer: (A) → (q); (B) → (s); (C) → (r); (D) → (p)