Question:

**Concept:** The number of triangles formed by joining vertices of an $n$-sided polygon is $T_n = \binom{n}{3}$. **Step 1:** Express $T_n$ using the combination formula. $$T_n = \binom{n}{3} = \frac{n(n-1)(n-2)}{6}$$ **Step 2:** Express $T_{n+1}$ using the combination formula. $$T_{n+1} = \binom{n+1}{3} = \frac{(n+1)n(n-1)}{6}$$ **Step 3:** Calculate the difference $T_{n+1} - T_n$. $$T_{n+1} - T_n = \frac{(n+1)n(n-1)}{6} - \frac{n(n-1)(n-2)}{6}$$ $$T_{n+1} - T_n = \frac{n(n-1)}{6} [(n+1) - (n-2)]$$ $$T_{n+1} - T_n = \frac{n(n-1)}{6} [n+1-n+2]$$ $$T_{n+1} - T_n = \frac{n(n-1)}{6} [3]$$ $$T_{n+1} - T_n = \frac{n(n-1)}{2}$$ **Step 4:** Apply the given condition $T_{n+1} - T_n = 10$. $$\frac{n(n-1)}{2} = 10$$ **Step 5:** Solve the resulting equation for $n$. $$n(n-1) = 20$$ $$n^2 - n - 20 = 0$$ Factor the quadratic equation: $$(n-5)(n+4) = 0$$ This yields two possible values for $n$: $n=5$ or $n=-4$. **Step 6:** Determine the valid value of $n$. Since $n$ represents the number of sides of a polygon, $n$ must be a positive integer. Furthermore, for $T_n = \binom{n}{3}$ to be defined, $n \ge 3$. Therefore, $n=5$ is the only valid solution.