Step 1: Understand the Formula for Tn
A regular polygon of n sides has n vertices. To form a single triangle, we need to choose any 3 distinct vertices out of these n vertices.
The number of ways to choose 3 vertices out of n is given by the combination formula nC3:
Tn=(3n)=3×2×1n(n−1)(n−2)
Similarly, for a polygon with n+1 sides, the number of triangles formed is:
Tn+1=(3n+1)=3×2×1(n+1)(n)(n−1)
Step 2: Set up the Given Equation
We are given the condition:
Tn+1−Tn=21
Substitute the values of Tn+1 and Tn into the equation:
(3n+1)−(3n)=21
Using the fundamental combination identity (rn+1)−(rn)=(r−1n), we can simplify this directly:
(2n)=21
(Alternatively, you can expand it algebraically):
6(n+1)n(n−1)−6n(n−1)(n−2)=21
Factor out the common term 6n(n−1):
6n(n−1)⋅[(n+1)−(n−2)]=21
6n(n−1)⋅[n+1−n+2]=21
6n(n−1)⋅3=21
2n(n−1)=21
Step 3: Solve for n
Multiply both sides by 2:
n(n−1)=42
Expand into a standard quadratic equation:
n2−n−42=0
Factorize the quadratic equation by splitting the middle term:
n2−7n+6n−42=0
n(n−7)+6(n−7)=0
(n−7)(n+6)=0
This gives two possible values for n:
n=7orn=−6
Since the number of sides of a polygon must be a positive integer, we discard n=−6. Therefore:
n=7
Correct Answer:
(b) 7
