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 [2001]

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 [2001]

Smart Achievers · Accepted Answer

(2)

$T_{n} = {}^{n}C_{3}; T_{n + 1} = {}^{n + 1}C_{3}$

$Now, T_{n + 1} - T_{n} = 21 \Rightarrow {}^{n + 1}C_{3} - {}^{n}C_{3} = 21$

$\Rightarrow \frac{(n + 1) n (n - 1)}{3 \cdot 2 \cdot 1} - \frac{n (n - 1) (n - 2)}{3 \cdot 2 \cdot 1} = 21$

$\Rightarrow n (n - 1) (n + 1 - (n - 2)) = 126$

$\Rightarrow n (n - 1) = 42$

$\Rightarrow n (n - 1) = 7 \times 6 \Rightarrow n = 7$

Solution

Q.
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 [2001]

1 5

2 7

3 6

4 4

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Solution

Q. Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon of n sides. If Tn+1-Tn=21, then n equals [2001]

1 5

2 7

3 6

4 4

Ans. (2) Tn=C3n; Tn+1=C3n+1 Now, Tn+1-Tn=21 ⇒ C3n+1-C3n=21 ⇒(n+1)n(n-1)3·2·1-n(n-1)(n-2)3·2·1=21 ⇒n(n-1)(n+1-(n-2))=126 ⇒n(n-1)=42 ⇒n(n-1)=7×6 ⇒ n=7

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Q.
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 [2001]