Binomial TheoremmediumPYQ · JEE Main · 4 Apr 2026 · Shift 1 (Morning)Free

Coefficient of x³ Condition: p + n = 11 | JEE Main 2026

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

Question

Let the smallest value of

k\in\mathbb{N}

, for which the coefficient of

x^3

(1+x)^3+(1+x)^4+(1+x)^5+\cdots+(1+x)^{99}+(1+kx)^{100}

x\ne0

, is

\left(43n+\dfrac{101}{4}\right)\big({}^{100}C_3\big)

for some

n\in\mathbb{N}

, be

p

. Then the value of

p+n

10

11

correct

12

13

Solution

Step 1: The coefficient of

x^3

from the sum

(1+x)^3+\cdots+(1+x)^{99}

is, by the hockey-stick identity,

{}^3C_3+{}^4C_3+\cdots+{}^{99}C_3={}^{100}C_4.

From

(1+kx)^{100}

the coefficient of

x^3

{}^{100}C_3\,k^3

. Step 2: So the total coefficient is

{}^{100}C_4+{}^{100}C_3\,k^3=\left(43n+\frac{101}{4}\right){}^{100}C_3.

Divide by

{}^{100}C_3

(using

\dfrac{{}^{100}C_4}{{}^{100}C_3}=\dfrac{97}{4}

\frac{97}{4}+k^3=43n+\frac{101}{4}.

Step 3: Thus

k^3=43n+1

. The smallest

k\in\mathbb{N}

making

43n+1

a perfect cube is

k^3=216

(i.e.

n=5

), so

k=6=p

. Step 4:

p+n=6+5=11.

Correct answer: (2)

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.