Binomial TheoremmediumPYQ · JEE Advanced · 2 Apr 2026 · Shift 2 (Afternoon)Free

Triangles in a Polygon: p(n+1) - p(n) = 66 gives Sum 5 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
Let pnp_n denote the total number of triangles formed by joining the vertices of an nn-side regular polygon. If pn+1−pn=66p_{n+1}-p_n=66, then the sum of all distinct prime divisors of nn is
A77
B88
C55correct
D66
Solution
Step 1: A triangle is formed by choosing 3 vertices, so
pn=nC3.p_n={}^nC_3.
 Step 2: Then
pn+1−pn=n+1C3−nC3=(n+1)n(n−1)6−n(n−1)(n−2)6=66.p_{n+1}-p_n={}^{n+1}C_3-{}^nC_3=\frac{(n+1)n(n-1)}{6}-\frac{n(n-1)(n-2)}{6}=66.
 Step 3: Factor n(n−1)6\dfrac{n(n-1)}{6}:
n(n−1)6[(n+1)−(n−2)]=n(n−1)6⋅3=66.\frac{n(n-1)}{6}\big[(n+1)-(n-2)\big]=\frac{n(n-1)}{6}\cdot 3=66.
 Step 4: Simplify:
n(n−1)2=66 ⇒ n(n−1)=132 ⇒ n=12.\frac{n(n-1)}{2}=66\ \Rightarrow\ n(n-1)=132\ \Rightarrow\ n=12.
 Step 5: Prime divisors of 1212 are 22 and 33, so their sum is 2+3=52+3=5. Correct answer: (3)
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