Aspire Faculty ID #10326 · Topic: NIMCET 2013 · Just now
NIMCET 2013

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

Solution

$T_{n+1} - T_n = 21$

$\binom{n+1}{3} - \binom{n}{3} = 21$

$\frac{(n+1)n(n-1)}{6} - \frac{n(n-1)(n-2)}{6} = 21$

$\frac{n(n-1)}{6}\left[(n+1)-(n-2)\right] = 21$

$\frac{n(n-1)}{6}\times 3 = 21$

$\frac{n(n-1)}{2} = 21$

$n(n-1) = 42$

$n^2 - n - 42 = 0$

$(n-7)(n+6) = 0$

$n = 7$

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