Question

Let $T_n$ denote the number of triangles which can be formed using the vertices of a regular polygon of $n$ sides. If $T_{n+1}-T_n=21$, then $n$ equals

• A. $5$
• B. $7$
• C. $6$
• D. $4$

Answer 1

Answer: (B)

$7$

Number of triangles formed using the vertices of a regular polygon of $n$ sides $=n_{C_3}$
Given, $T_{n+1}-T_n=21$
$=>{(n+1)}_{C_3}-n_{C_3}=21$
$=>n_{C_2}+n_{C_3}-n_{C_3}-21$ Since, ${(n+1)}_{C_r}=n_{C_{r-1}}+n_{C_r}$
$=>\frac{n!}{2!(n-2)!}=2$ Since (${n_C}_r=\frac{n!}{r!(n-r)!}$)
$=>\frac{n\times (n-1)\times (n-2)!}{2\times (n-2)!}=21$
$=>n^2-n=42$
$=>n^2-n-42=0$
$=>n^2-7n+6n-42=0$
$=>n(n-7)+6(n-7)=0$
$=>n=7$ or $-6$
Since, $n$ cannot be negative, $n=7$