Suppose t_n is the number of triangles formed | vedclass

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Question

(C) The number of triangles formed by choosing $3$ vertices from $n$ vertices of a regular polygon is given by $t_n = \binom{n}{3} = \frac{n(n-1)(n-2)

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) The number of triangles formed by choosing $3$ vertices from $n$ vertices of a regular polygon is given by $t_n = \binom{n}{3} = \frac{n(n-1)(n-2)}{6}$.Given the condition $t_{n+1} = t_n + 28$,we substitute the formula:$\binom{n+1}{3} - \binom{n}{3} = 28$.Using the property $\binom{n+1}{3} - \binom{n}{3} = \binom{n}{2}$,we get:$\binom{n}{2} = 28$.$\frac{n(n-1)}{2} = 28$.$n(n-1) = 56$.$n^2 - n - 56 = 0$.$(n-8)(n+7) = 0$.Since $n$ must be positive,$n = 8$.

Suppose $t_n$ is the number of triangles formed using the vertices of a regular polygon of $n$ sides. If $t_{n+1} = t_n + 28$,then $n =$

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