Let the smallest value of k in mathbb N , for which the coefficient of x 3 in (1+x) 3 + (1+x) 4 + (1+x) 5 + ldots + (1+x) 99 + (1+kx) 100 is (101/4+43n )binom 100 3 for some ninmathbb N , be p . Then the value of p+n is:

Mathematics · Binomial Theorem

JEE Main 2026 — 4 April, Morning Shift — Question 32

Let the smallest value of $k \in \mathbb{N}$ , for which the coefficient of $x^3$ in $(1+x)^3 + (1+x)^4 + (1+x)^5 + \ldots + (1+x)^{99} + (1+kx)^{100}$ is $\left(\frac{101}{4}+43n\right)\binom{100}{3}$ for some $n\in\mathbb{N}$ , be $p$ . Then the value of $p+n$ is:

Option A:
10
Option B:
11
Correct
Option C:
12
Option D:
13

Answer: B

Step-by-step solution

${ }^{3} \mathrm{C}_{3}+{ }^{4} \mathrm{C}_{3}+{ }^{5} \mathrm{C}_{3}+\ldots+{ }^{99} \mathrm{C}_{3}+{ }^{100} \mathrm{C}_{3} \mathrm{~K}^{3}=\left(43 \mathrm{n}+\frac{101}{4}\right){ }^{100} \mathrm{C}_{3}$

\\& \mathrm{n}=5 \\& \mathrm{~K}^{3}=216 \\& \mathrm{~K}=6=\mathrm{p} \\& \mathrm{p}+\mathrm{n}=11\end{aligned}$$

Answer key and solution verified before publishing.

Practise Binomial Theorem

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 4 April, Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Introduction to Binomial Theorem