Binomial Theorem
Pascal's identity / combinatorial identities
Grade 11
Question:
<p>We have \({}^nC_{r+1} + {}^nC_{r-1} + 2 \times {}^nC_r\). Which of the following is equal to this expression?</p>
<p>\({}^{n+1}C_{r+1}\)</p>
<p>\({}^{n+2}C_{r+1}\)</p>
<p>\({}^{n+2}C_r\)</p>
<p>\({}^{n+1}C_r\)</p>
Step-by-Step Solution
Key Concept: Recognize that the expression can be rewritten by combining binomial coefficients strategically: use the identity ${}^nC_r + {}^nC_{r+1} = {}^{n+1}C_{r+1}$ twice to telescope the sum into a single binomial coefficient.
<p><strong>Step 1:</strong> Rewrite the expression by grouping strategically:</p><p>${}^nC_{r+1} + {}^nC_{r-1} + 2 \times {}^nC_r = ({}^nC_{r-1} + {}^nC_r) + ({}^nC_r + {}^nC_{r+1})$</p><p><strong>Step 2:</strong> Apply Pascal's identity ${}^nC_r + {}^nC_{r+1} = {}^{n+1}C_{r+1}$ to each group:</p><p>$({}^nC_{r-1} + {}^nC_r) + ({}^nC_r + {}^nC_{r+1}) = {}^{n+1}C_r + {}^{n+1}C_{r+1}$</p><p><strong>Step 3:</strong> Apply Pascal's identity once more:</p><p>${}^{n+1}C_r + {}^{n+1}C_{r+1} = {}^{n+2}C_{r+1}$</p><p>∴ Answer: B (which is ${}^{n+2}C_{r+1}$)</p>
Correct Answer: B