Let `T_n` donote the number of traingles which can be formed using the vertices of a regular polygon of n sides. If `T_(n+1) -T_n=21` then n equals

Options

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

Answer

To solve the problem, we need to determine the value of \( n \) such that the difference in the number of triangles that can be formed using the vertices of a regular polygon with \( n \) sides and \( n+1 \) sides is equal to 21.

1. **Understanding the number of triangles**:
The number of triangles \( T_n \) that can be formed from the vertices of a polygon with \( n \) sides is given by the combination formula:
\[ T_n = \binom{n}{3} = \frac{n(n-1)(n-2)}{6} \]
Similarly, for \( n+1 \):
\[ T_{n+1} = \binom{n+1}{3} = \frac{(n+1)n(n-1)}{6} \]

2. **Setting up the equation**:
We know that:
\[ T_{n+1} - T_n = 21 \]
Substituting the formulas we derived:
\[ \frac{(n+1)n(n-1)}{6} - \frac{n(n-1)(n-2)}{6} = 21 \]

3. **Simplifying the equation**:
Combine the fractions:
\[ \frac{(n+1)n(n-1) - n(n-1)(n-2)}{6} = 21 \]
Multiply both sides by 6 to eliminate the fraction:
\[ (n+1)n(n-1) - n(n-1)(n-2) = 126 \]

4. **Factoring out common terms**:
Notice that \( n(n-1) \) is common in both terms:
\[ n(n-1)((n+1) - (n-2)) = 126 \]
Simplifying the expression inside the parentheses:
\[ n(n-1)(3) = 126 \]
Therefore:
\[ n(n-1) = 42 \]

5. **Forming a quadratic equation**:
Rearranging gives us:
\[ n^2 - n - 42 = 0 \]

6. **Factoring the quadratic**:
To factor \( n^2 - n - 42 \):
\[ (n - 7)(n + 6) = 0 \]
This gives us two potential solutions:
\[ n - 7 = 0 \quad \text{or} \quad n + 6 = 0 \]
Thus:
\[ n = 7 \quad \text{or} \quad n = -6 \]

7. **Choosing the valid solution**:
Since \( n \) must be a positive integer (as it represents the number of sides of a polygon), we discard \( n = -6 \).

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