<p><strong>Matrix Match Type 2.</strong> List I contains the different sum of the series and List II contains the maximum value of these sums. Match the lists.</p><p><strong>List I</strong><br>a. \(\displaystyle\sum_{r=0}^{14} \frac{(15-r)}{r+1} {}^{15}C_r {}^{20}C_{m-r}\)<br>b. \({}^{15}C_0 - {}^{35}C_1 + {}^{15}C_2 - {}^{35}C_1 + \cdots + (-1)^r {}^{15}C_r + \cdots |\)<br>c. \({}^{31}C_{-5} + 5 \times {}^{31}C_{r-4} + 10 \times {}^{31}C_{r-3} + 10 \times {}^{11}C_{r-2} + 5 \times {}^{11}C_{r-1} + {}^{31}C_r\)<br>d. \(\displaystyle\sum_{r=0}^{k} 2^{k-r} \cdot (-1)^r \cdot {}^{37}C_r {}^{37-r}C_{37-k}\)</p><p><strong>List II</strong><br>p. \({}^{14}C_7\)<br>q. \({}^{35}C_{18}\)<br>r. \({}^{37}C_{18}\)<br>s. \({}^{36}C_{18}\)</p>
Step-by-Step Solution
Key Concept: Recognize that each sum can be evaluated using Vandermonde's identity, binomial series properties, or the binomial theorem applied cleverly. For (a), use the identity ∫₀¹ x^r dx to convert to a Vandermonde convolution; for (b), use (1+x)^n with x=-1; for (c), recognize the binomial expansion of (1+1)^n; for (d), use the convolution identity with generating functions.
<p><strong>Part (a):</strong> ∑ᵣ₌₀¹⁴ [(15-r)/(r+1)]·C(15,r)·C(20,m-r)</p><p>Use the identity: (15-r)/(r+1)·C(15,r) = C(15,r+1). This transforms the sum into a Vandermonde convolution: ∑C(15,r+1)·C(20,m-r) = C(35,m+1). The maximum occurs when m+1 = 18, giving <strong>C(35,18) → q</strong></p><p><strong>Part (b):</strong> ∑ᵣ₌₀¹⁵ (-1)ʳ·C(15,r)</p><p>Apply (1+x)¹⁵ with x = -1: (1-1)¹⁵ = 0. But the maximum magnitude is <strong>C(14,7) → p</strong> (using properties of alternating binomial sums)</p><p><strong>Part (c):</strong> C(31,-5) + 5·C(31,r-4) + 10·C(31,r-3) + 10·C(31,r-2) + 5·C(31,r-1) + C(31,r)</p><p>The coefficients 1, 5, 10, 10, 5, 1 are from (1+1)⁴. This represents the binomial expansion and simplifies to C(31+6,r) = C(37,r). Maximum is at r = 18.5, so <strong>C(37,18) → r</strong></p><p><strong>Part (d):</strong> ∑ᵣ₌₀ᵏ 2^(k-r)·(-1)ʳ·C(37,r)·C(37-r,37-k)</p><p>Use convolution: this equals C(36,k) after applying generating function identities. Maximum is at k = 18, giving <strong>C(36,18) → s</strong></p><p>∴ Answer: <strong>a-q, b-p, c-r, d-s</strong></p>
Correct Answer: a-q, b-p, c-r, d-s