Let $p_{n}$ denote the total number of triangles formed by joining the vertices of an n -side regula

Question

Let $p_{n}$ denote the total number of triangles formed by joining the vertices of an n -side regular polygon. If $p_{n + 1} - p_{n} = 66$, then the sum of all distinct prime divisors of n is :

Accepted Answer

Sol. [Key Concept: $p_{n} = {}^{n}C_{3}$; solve $p_{n + 1} - p_{n} = 66$ for $n$, then sum its prime divisors.]$${P_{n} = \ ^{n}C_{3} }{P_{n + 1} - P_{n} = 66 }{\ ^{n + 1}C_{3} - \ ^{n}C_{3} = 66 }{\Rightarrow \frac{(n + 1)n(n - 1)}{6} - \frac{n(n - 1)(n - 2)}{6} = 66 }{\Rightarrow \frac{n(n - 1)}{6}\lbrack n + 1 - n + 2\rbrack = 66 }{\Rightarrow n(n - 1) = 132 }{n = 12 }$$Prime divisors of 12 are 2 & 3sum $= 2 + 3 = 5$

Question

Options

Solution

More Progression (Sequence and Series) Questions