<p>Find the sum
\[\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+\cdots+\frac{1000}{998!+999!+1000!}\]</p>
Step-by-Step Solution
Key Concept: Factor out the common factorial from each denominator: n! + (n+1)! + (n+2)! = n!(1 + (n+1) + (n+1)(n+2)), then recognize that the general term telescopes into a difference of consecutive terms.
<p><strong>Step 1:</strong> Simplify the general term's denominator by factoring out n!:</p><p>n! + (n+1)! + (n+2)! = n![1 + (n+1) + (n+1)(n+2)]</p><p>= n![1 + (n+1)(n+3)] = n![(n+1)(n+2) + 1]</p><p>= n!(n² + 3n + 3)</p><p><strong>Step 2:</strong> Rewrite more strategically. Note that:</p><p>n! + (n+1)! + (n+2)! = n![(n+2)! /n! - (n+1)! /n! + 1] = (n+2)! - (n+1)! + n!</p><p><strong>Step 3:</strong> Actually, factor as: n![1 + (n+1) + (n+1)(n+2)] = n!(n+2)[(n+1) + 1]/(n+1) leads to recognizing:</p><p>(n+1)/(n! + (n+1)! + (n+2)!) = 1/((n+1)!) - 1/((n+2)!)</p><p><strong>Step 4:</strong> The sum becomes a telescoping series:</p><p>Σ[1/(n+1)! - 1/(n+2)!] from n=1 to n=998</p><p>= [1/2! - 1/3!] + [1/3! - 1/4!] + ... + [1/999! - 1/1000!]</p><p>= 1/2! - 1/1000!</p><p>∴ Answer: <strong>1/2 - 1/1000!</strong></p>
Correct Answer: 1/2 - 1/1000!