<p>Find the sum to <em>n</em> terms of the series <br>\(1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + \cdots\)</p>
Step-by-Step Solution
Key Concept: The general term is r(r+1)(r+2), which can be decomposed using the telescoping identity: r(r+1)(r+2) = ¼[r(r+1)(r+2)(r+3) - (r-1)r(r+1)(r+2)]. Summing this telescoping series leaves only the last term.
<p><strong>Step 1:</strong> Identify the general term as T<sub>r</sub> = r(r+1)(r+2)</p><p><strong>Step 2:</strong> Use the telescoping identity: r(r+1)(r+2) = ¼[r(r+1)(r+2)(r+3) - (r-1)r(r+1)(r+2)]</p><p><strong>Verification:</strong> ¼[r(r+1)(r+2)(r+3) - (r-1)r(r+1)(r+2)] = ¼·r(r+1)(r+2)[(r+3) - (r-1)] = ¼·r(r+1)(r+2)·4 = r(r+1)(r+2) ✓</p><p><strong>Step 3:</strong> Sum from r=1 to n: S<sub>n</sub> = ¼∑[r(r+1)(r+2)(r+3) - (r-1)r(r+1)(r+2)]</p><p><strong>Step 4:</strong> This is a telescoping series. Writing out terms:</p><p>r=1: ¼[1·2·3·4 - 0]</p><p>r=2: ¼[2·3·4·5 - 1·2·3·4]</p><p>r=3: ¼[3·4·5·6 - 2·3·4·5]</p><p>⋮</p><p>r=n: ¼[n(n+1)(n+2)(n+3) - (n-1)n(n+1)(n+2)]</p><p><strong>Step 5:</strong> All intermediate terms cancel. Only the final term survives: S<sub>n</sub> = ¼·n(n+1)(n+2)(n+3)</p><p><strong>∴ Answer:</strong> <strong>n(n+1)(n+2)(n+3)/4</strong></p>
Correct Answer: n(n+1)(n+2)(n+3)/4