Let `T_n ` denote the number of triangles which can be be formed using the vertices of a regular polygon of n sides. If ` T_(n+1) - T _n = 21, " then " n = `

Options

A. 5
B. 7
C. 6
D. 4

Answer

To solve the problem, we need to find the value of \( n \) such that \( T_{n+1} - T_n = 21 \), where \( T_n \) represents the number of triangles that can be formed using the vertices of a regular polygon with \( n \) sides.

### Step-by-Step Solution:

1. **Understanding \( T_n \)**:
The number of triangles that can be formed using the vertices of a polygon with \( n \) sides is given by the combination formula \( T_n = \binom{n}{3} \). This is because we need to choose 3 vertices from \( n \) vertices to form a triangle.

2. **Expressing \( T_{n+1} \)**:
Similarly, for \( n+1 \) sides, the number of triangles is:
\[ T_{n+1} = \binom{n+1}{3} \]

3. **Setting up the equation**:
We are given that:
\[ T_{n+1} - T_n = 21 \]
Substituting the expressions for \( T_{n+1} \) and \( T_n \):
\[ \binom{n+1}{3} - \binom{n}{3} = 21 \]

4. **Using the combination formula**:
The combination formula is:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Therefore,
\[ \binom{n+1}{3} = \frac{(n+1)n(n-1)}{6} \]
and
\[ \binom{n}{3} = \frac{n(n-1)(n-2)}{6} \]

5. **Substituting into the equation**:
Now substituting these into our equation:
\[ \frac{(n+1)n(n-1)}{6} - \frac{n(n-1)(n-2)}{6} = 21 \]

6. **Simplifying the equation**:
Factor out \( \frac{n(n-1)}{6} \):
\[ \frac{n(n-1)}{6} \left( (n+1) - (n-2) \right) = 21 \]
Simplifying the term inside the parentheses:
\[ (n+1) - (n-2) = 3 \]
Thus, we have:
\[ \frac{n(n-1) \cdot 3}{6} = 21 \]
Simplifying further:
\[ \frac{n(n-1)}{2} = 21 \]

7. **Multiplying both sides by 2**:
\[ n(n-1) = 42 \]

8. **Rearranging into a standard quadratic equation**:
\[ n^2 - n - 42 = 0 \]

9. **Factoring the quadratic**:
To factor \( n^2 - n - 42 \), we look for two numbers that multiply to \(-42\) and add to \(-1\). These numbers are \( -7 \) and \( 6 \):
\[ (n - 7)(n + 6) = 0 \]

10. **Finding the values of \( n \)**:
Setting each factor to zero gives:
\[ n - 7 = 0 \quad \Rightarrow \quad n = 7 \]
\[ n + 6 = 0 \quad \Rightarrow \quad n = -6 \quad (\text{not valid since } n \text{ must be positive}) \]

Thus, the solution is:
\[ \boxed{7} \]