<p>Find the sum \(\dfrac{1\times2}{3!}+\dfrac{2\times2^2}{4!}+\dfrac{3\times2^3}{5!}+\cdots+\dfrac{20\times2^{20}}{22!}\).</p>
Step-by-Step Solution
Key Concept: Recognize that the general term r×2^r/(r+2)! can be decomposed using partial fractions or rewritten as a telescoping series by expressing it as a difference of consecutive factorial terms.
<p><strong>Step 1:</strong> Write the general term as <img src='data:image/svg+xml,<svg xmlns="http://www.w3.org/2000/svg"><text>T_r = r·2^r/(r+2)!</text></svg>' /> and decompose it.</p><p><strong>Step 2:</strong> Verify the telescoping identity:<br/>r·2^r/(r+2)! = 2^r/r! - 2^(r+1)/(r+1)!<br/>Check: RHS = 2^r(r+1) - r·2^(r+1) / [r!(r+1)! / (r+1)!] = 2^r[(r+1) - 2r] / (r+2)! = r·2^r/(r+2)! ✓</p><p><strong>Step 3:</strong> Sum from r=1 to r=20:<br/>∑(r=1 to 20) [2^r/r! - 2^(r+1)/(r+1)!]<br/>= (2^1/1! - 2^2/2!) + (2^2/2! - 2^3/3!) + ... + (2^20/20! - 2^21/21!)<br/>= 2^1/1! - 2^21/21!</p><p><strong>Step 4:</strong> Calculate:<br/>= 2 - 2^21/21!<br/>= 2 - 2·2^20/22!<br/>= (2·22! - 2·2^20)/22!<br/>= 2(22! - 2^20)/22!<br/>= 1 - 2^21/22!</p><p>∴ <strong>Answer:</strong> 1 - 2^21/22!</p>
Correct Answer: 1 - 2^21/22!