Question

Let ${{T}_{n}}$ be the number of triangles of which can be formed using the vertices of a regular polygon of n sides. If ${{T}_{n+1}}+{{T}_{n}}=91$, find the value of n.
A. 8
B. 7
C. 13
D. 9

Answer 1

Hint: We first describe how triangles are formed and how we can make triangles by joining vertices of a n-sided polygon. Then we find the formula of choosing r things out of n things. We use the formula to find the number of ways 3 points can be chosen from the 10 points. We put the values in the equation and solve it to find the solution of the problem.

Complete step by step answer:
A n-sided polygon has n vertices and n sides.
We know that to form a triangle we need three points which are non-linear.
We also have that no three points out of those n vertices are non-linear.
As we are trying to create triangles by joining the vertices of a n-sided polygon, we have to choose 3 points out of those n points.
${{T}_{n}}$ be the number of triangles of which can be formed using the vertices of a regular polygon of n sides.
So, the number of triangles created by joining the vertices of a n-sided polygon is exactly equal to the number of ways 3 points can be chosen from the n points.
We know the number of ways r things can be chosen out of n things are ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}$.
So, the number of ways 3 points can be chosen from the n points is ${{T}_{n}}={}^{n}{{C}_{3}}$.
It’s given that ${{T}_{n+1}}+{{T}_{n}}=91$. We put the value and get
$\begin{align}
  & {{T}_{n+1}}+{{T}_{n}}=91 \\
 & \Rightarrow {}^{n+1}{{C}_{3}}+{}^{n}{{C}_{3}}=91 \\
 & \Rightarrow \dfrac{\left( n+1 \right)!}{\left( n+1-3 \right)!\times 3!}+\dfrac{n!}{\left( n-3 \right)!\times 3!}=91 \\
\end{align}$
We now the form of factorial to find solution
$\begin{align}
  & \dfrac{\left( n+1 \right)!}{\left( n-2 \right)!\times 3!}+\dfrac{n!}{\left( n-3 \right)!\times 3!}=91 \\
 & \Rightarrow n\left( n+1 \right)\left( n-1 \right)+n\left( n-1 \right)\left( n-2 \right)=91\times 6 \\
 & \Rightarrow n\left( n-1 \right)\left( 2n-1 \right)=546 \\
 & \Rightarrow 2{{n}^{3}}-3{{n}^{2}}+n-546=0 \\
 & \Rightarrow \left( n-7 \right)\left( 2{{n}^{2}}+11n+78 \right)=0 \\
\end{align}$
Solving the equation, we get $n=7$ and the other two roots will be imaginary.

So, the correct answer is “Option B”.

Note: We have to remember that in case of triangles we don’t have to worry but if we are asked about diagonal where we need any 2 points then at the end, we have to subtract the number of sides from the option as they will be considered as diagonals. So, the answer will be ${}^{n}{{C}_{2}}-n$.