<p>Let <em>T<sub>n</sub></em> denote the number of triangles which can be formed using the vertices of a regular polygon on <em>n</em> sides. If \(T_{n-1} - T_n = 21\), then <em>n</em> equals</p>
Step-by-Step Solution
Key Concept: The number of triangles from n vertices is C(n,3). Set up T_{n-1} - T_n = C(n-1,3) - C(n,3) = 21 and simplify using the combinatorial identity to find n directly.
<p><strong>Step 1:</strong> Express T_n in terms of combinations. The number of triangles formed by n vertices of a regular polygon is: T_n = C(n,3) = n(n-1)(n-2)/6</p><p><strong>Step 2:</strong> Write the given condition: T_{n-1} - T_n = 21</p><p>C(n-1,3) - C(n,3) = 21</p><p><strong>Step 3:</strong> Expand using the combinatorial formula:</p><p>(n-1)(n-2)(n-3)/6 - n(n-1)(n-2)/6 = 21</p><p><strong>Step 4:</strong> Factor out common terms:</p><p>(n-1)(n-2)/6 [(n-3) - n] = 21</p><p>(n-1)(n-2)/6 × (-3) = 21</p><p>-(n-1)(n-2)/2 = 21</p><p>(n-1)(n-2) = -42</p><p><strong>Step 5:</strong> This gives a negative product, so reconsider: T_n - T_{n-1} = 21 is the intended form.</p><p>n(n-1)(n-2)/6 - (n-1)(n-2)(n-3)/6 = 21</p><p>(n-1)(n-2)/6 [n - (n-3)] = 21</p><p>(n-1)(n-2)/6 × 3 = 21</p><p>(n-1)(n-2)/2 = 21</p><p>(n-1)(n-2) = 42</p><p><strong>Step 6:</strong> Solve: n² - 3n + 2 = 42 → n² - 3n - 40 = 0</p><p>(n-8)(n+5) = 0</p><p>Since n > 0, n = 8</p><p>∴ Answer: B</p>
Correct Answer: B