<p>The value of \({}^nC_1 + {}^{n+1}C_2 + {}^{n+2}C_3 + \ldots + {}^{n+m-1}C_m\) is equal to</p>
<p>(1) \({}^{m+n}C_{n-1}\)</p>
<p>(2) \({}^{m+n}C_{n-1}\)</p>
<p>(3) \({}^mC_1 + {}^{m+1}C_2 + {}^{m+2}C_3 + \ldots + {}^{m+n-1}C_n\)</p>
<p>(4) \({}^{m+n}C_{m} - 1\)</p>
Step-by-Step Solution
Key Concept: Use the hockey stick identity: ∑(r=1 to m) C(n+r-1, r) = C(n+m, m) - 1, which collapses the sum by recognizing the telescoping pattern in binomial coefficients.
<p><strong>Step 1:</strong> Recognize the sum has the form ∑(k=1 to m) C(n+k-1, k)</p><p><strong>Step 2:</strong> Apply the Hockey Stick Identity: ∑(k=1 to m) C(n+k-1, k) = ∑(k=0 to m) C(n+k, k) - C(n, 0)</p><p><strong>Step 3:</strong> The Hockey Stick Identity states: ∑(k=0 to m) C(n+k, k) = C(n+m+1, m)</p><p><strong>Step 4:</strong> Therefore: C(n+1)C(2) + C(n+2)C(3) + ... + C(n+m-1)C(m) = C(n+m+1, m) - 1</p><p><strong>Step 5:</strong> Simplify: This equals <strong>C(n+m, m) - 1</strong> or equivalently <strong>C(n+m+1, m) - 1</strong> depending on indexing</p><p>∴ Answer: <strong>C(n+m, m) - 1</strong> or <strong>C(n+m+1, m) - 1</strong></p>
Correct Answer: C