The sum of the coefficients of x^ 499 and x^ 500 in (1+x)^ 1000 +x(1+x)^ 999 +x^ 2 (1+x)^ 998 + +x^ 1000 is :

Correct answer: A. ^ 1002 C_ 500 S = (1+x)^ 1000 + x(1+x)^ 999 + x^2(1+x)^ 998 + + x^ 1000 This is a geometric series with first term (1+x)^ 1000 , common ratio (x)/(1+x) , and 1001 terms. S = (1+x)^ 1000 · 1 - ((x)/(1+x))^ 1001 1 - (x)/(1+x) = (1+x)^ 1001 - x^ 1001 Required sum = coefficient of x^ 499 + coefficient of x^ 500 in (1+x)^ 1001 - x^ 1001 = ^ 1001 C_ 499 + ^ 1001 C_ 500 = ^ 1002 C_ 500

JEE Main 202628 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 28 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2026 · 28 January · Evening Shift · Mathematics · Binomial Theorem

The sum of the coefficients of x^ 499 and x^ 500 in (1+x)^ 1000...

The sum of the coefficients of

x^{499}

and

x^{500}

(1+x)^{1000}+x(1+x)^{999}+x^{2}(1+x)^{998}+\ldots+x^{1000}

is :

Options

A ${ }^{1002} C_{500}$
B ${ }^{1002} C_{501}$
C ${ }^{1001} C_{501}$
D ${ }^{1000} C_{501}$

Correct answer

A. ${ }^{1002} C_{500}$

Step-by-step solution

S = (1+x)^{1000} + x(1+x)^{999} + x^2(1+x)^{998} + \cdots + x^{1000}

This is a geometric series with first term

(1+x)^{1000}

, common ratio

\frac{x}{1+x}

, and 1001 terms.

S = (1+x)^{1000} \cdot \dfrac{1 - \left(\frac{x}{1+x}\right)^{1001}}{1 - \frac{x}{1+x}}

= (1+x)^{1001} - x^{1001}

Required sum = coefficient of

x^{499}

+ coefficient of

x^{500}

(1+x)^{1001} - x^{1001}

= {}^{1001}C_{499} + {}^{1001}C_{500} = {}^{1002}C_{500}

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