Let `T_n` be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If `T_(n+1)-T_n=""10` , then the value of n is (1) 5 (2) 10 (3) 8 (4) 7

Options

A. 5
B. 10
C. 8
D. 7

Explanation

To solve the problem, we need to find the value of \( n \) such that the difference in the number of triangles formed by \( n \)-sided and \( (n+1) \)-sided regular polygons equals 10.

### Step-by-Step Solution:

1. **Understanding the Number of Triangles**:
The number of triangles that can be formed from the vertices of an \( n \)-sided polygon is given by the combination formula \( T_n = \binom{n}{3} \).

2. **Setting Up the Equation**:
We are given that:
\[ T_{n+1} - T_n = 10 \]
Substituting the formula for \( T_n \):
\[ \binom{n+1}{3} - \binom{n}{3} = 10 \]

3. **Using the Combination Property**:
We can use the property of combinations:
\[ \binom{n+1}{r} - \binom{n}{r} = \binom{n}{r-1} \]
Here, we set \( r = 3 \):
\[ \binom{n+1}{3} - \binom{n}{3} = \binom{n}{2} \]
Therefore, we have:
\[ \binom{n}{2} = 10 \]

4. **Calculating \( \binom{n}{2} \)**:
The formula for \( \binom{n}{2} \) is:
\[ \binom{n}{2} = \frac{n(n-1)}{2} \]
Setting this equal to 10:
\[ \frac{n(n-1)}{2} = 10 \]

5. **Solving for \( n \)**:
Multiply both sides by 2:
\[ n(n-1) = 20 \]
Rearranging gives us:
\[ n^2 - n - 20 = 0 \]

6. **Factoring the Quadratic Equation**:
We can factor the quadratic:
\[ (n - 5)(n + 4) = 0 \]
This gives us two potential solutions:
\[ n - 5 = 0 \quad \Rightarrow \quad n = 5 \]
\[ n + 4 = 0 \quad \Rightarrow \quad n = -4 \]

7. **Selecting the Valid Solution**:
Since \( n \) must be a positive integer (as it represents the number of sides of a polygon), we discard \( n = -4 \). Thus, we have:
\[ n = 5 \]

### Final Answer:
The value of \( n \) is \( 5 \).