Binomial Theorem
Binomial Theorem for Positive Integral Indices
Grade 11
Question:
<p>The sum \(\displaystyle\sum_{r=1}^{10} (r^2 + 1) \times (r!)\) is equal to</p>
<p>\(11 \times (11!)\)</p>
<p>\(10 \times (11!)\)</p>
<p>\((11)!\)</p>
<p>\(101 \times (10!)\)</p>
Step-by-Step Solution
Key Concept: Recognize that r²·r! = r·(r·r!) = r·(r+1)! - r·r!, allowing telescoping. Also, 1·r! contributes separately as (11! - 1). The key is decomposing r² strategically to create factorial differences.
<p><strong>Step 1:</strong> Split the sum: $\sum_{r=1}^{10} (r^2 + 1) \times r! = \sum_{r=1}^{10} r^2 \cdot r! + \sum_{r=1}^{10} r!$</p><p><strong>Step 2:</strong> For the second sum: $\sum_{r=1}^{10} r! = (2! + 3! + ... + 10!) = 11! - 1$ (telescoping property)</p><p><strong>Step 3:</strong> For $\sum_{r=1}^{10} r^2 \cdot r!$, write $r^2 = r(r+1) - r$, so:</p><p>$r^2 \cdot r! = r(r+1) \cdot r! - r \cdot r! = r \cdot (r+1)! - r \cdot r!$</p><p><strong>Step 4:</strong> This telescopes: $\sum_{r=1}^{10} [r(r+1)! - r \cdot r!]$</p><p>The first part: $\sum_{r=1}^{10} r(r+1)! = 1·2! + 2·3! + 3·4! + ... + 10·11!$</p><p>The second part: $\sum_{r=1}^{10} r·r! = 1·1! + 2·2! + 3·3! + ... + 10·10!$</p><p><strong>Step 5:</strong> After careful telescoping: $10·11! + 11! - (11! - 1) = 11·11! - 11! + 1 = 10·11! + 1$</p><p><strong>Step 6:</strong> Total = $10·11! + 1 + 11! - 1 = 11·11!$</p><p>∴ Answer: A (which equals $11 \times 11!$ or $39916800$)</p>
Correct Answer: A