<p>\((n+2)\,{}^nC_0 \cdot 2^{n-1} - (n+1)\,{}^nC_1 \cdot 2^n + n\,{}^nC_2 \cdot 2^{n-1} - \cdots\) is equal to</p>
Step-by-Step Solution
Key Concept: Recognize this as a derivative or weighted sum of binomial expansion terms. The coefficients (n+2), (n+1), n suggest applying the identity r·nCr = n·(n-1)C(r-1) or differentiating (1+x)^n and evaluating at specific x values.
<p><strong>Step 1:</strong> Rewrite the general term. The k-th term is (n+2-k)·<sup>n</sup>C<sub>k</sub>·2<sup>n-k</sup>·(-1)<sup>k</sup></p><p><strong>Step 2:</strong> Factor out 2<sup>n-1</sup>: The sum becomes 2<sup>n-1</sup>·Σ(n+2-k)·<sup>n</sup>C<sub>k</sub>·2<sup>1-k</sup>·(-1)<sup>k</sup></p><p><strong>Step 3:</strong> Split into two sums: (n+2)Σ<sup>n</sup>C<sub>k</sub>·2<sup>1-k</sup>·(-1)<sup>k</sup> - 2·Σk·<sup>n</sup>C<sub>k</sub>·2<sup>1-k</sup>·(-1)<sup>k</sup></p><p><strong>Step 4:</strong> For the first sum, use binomial expansion with x=-1/2: Σ<sup>n</sup>C<sub>k</sub>·2<sup>1-k</sup>·(-1)<sup>k</sup> = 2·(1-1/2)<sup>n</sup> = 2·(1/2)<sup>n</sup> = 2/2<sup>n</sup></p><p><strong>Step 5:</strong> For the second sum, use k·<sup>n</sup>C<sub>k</sub> = n·<sup>n-1</sup>C<sub>k-1</sub> and differentiation of binomial series to get: 2·n·(1/2)<sup>n-1</sup></p><p><strong>Step 6:</strong> Combine: 2<sup>n-1</sup>·[(n+2)·2/2<sup>n</sup> - 2·n·2/2<sup>n</sup>] = 2<sup>n-1</sup>·[2(n+2-2n)/2<sup>n</sup>] = 2<sup>n-1</sup>·2(2-n)/2<sup>n</sup></p><p>∴ Answer: <strong>1</strong> or <strong>2-n</strong> (depending on options provided)</p>
Correct Answer: A