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Question: Let \({T_n}\) denote the number of triangles which can be formed using the vertices of a regular pol...

Let Tn{T_n} denote the number of triangles which can be formed using the vertices of a regular polygon of n sides. If Tn+1−Tn=21{T_{n + 1}} - {T_n} = 21 then n equals
A. 5{\text{A}}{\text{. 5}}
B. 7{\text{B}}{\text{. 7}}
C. 6{\text{C}}{\text{. 6}}
D. 4{\text{D}}{\text{. 4}}

Explanation

Solution

Hint: - Here we choose three sides from n sides of a polygon by method of selection to form the triangle i.e.nC3{}^n{C_3}.Then similarly do for the n+1 sides of the polygon. After that apply the condition of the question.

Complete step by step solution:
Given that,
Tn+1−Tn=21{T_{n + 1}} - {T_n} = 21
Tn+1{T_{n + 1}} Can be written as n+1C3{}^{n + 1}{C_3}
⇒(n+1)C3−nC3=21\Rightarrow {}^{(n + 1)}{C_3} - {}^n{C_3} = 21
⇒nC2+nC3−nC3=21\Rightarrow {}^n{C_2} + {}^n{C_3} - {}^n{C_3} = 21
∵\because We know that (n+1)Cr=nCr−1+nCr{}^{(n + 1)}{C_r} = {}^n{C_{r - 1}} + {}^n{C_r}
⇒n!2!(n−2)!=21\Rightarrow \dfrac{{n!}}{{2!\left( {n - 2} \right)!}} = 21
∵\because We know that nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}
⇒n×(n−1)×(n−2)!2×(n−2)!=21\Rightarrow \dfrac{{n \times (n - 1) \times (n - 2)!}}{{2 \times (n - 2)!}} = 21
⇒n2−n=42 ⇒n2−n−42=0 ⇒n2−7n+6n−42=0 ⇒n(n−7)+6(n−7)=0 ⇒(n−7)(n+6)=0  \Rightarrow {n^2} - n = 42 \\\ \Rightarrow {n^2} - n - 42 = 0 \\\ \Rightarrow {n^2} - 7n + 6n - 42 = 0 \\\ \Rightarrow n(n - 7) + 6(n - 7) = 0 \\\ \Rightarrow (n - 7)(n + 6) = 0 \\\
∴\therefore n=7 0r n=-6
We know that sides cannot be negative ∴n=7\therefore n = 7 is the required answer.
Hence, option B is the correct answer.

Note:-Whenever we face such a type of question the key concepts for solving the question is that you have to first choose the three sides from the n sides by selection method to form the triangle and then proceed according to the condition which is given in the question.