Question
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 :
JEE Main
Mathematics
Binomial Theorem
2026
JEE Main 2026 (28 January Shift 2)
JEE Main Mathematics Question (2026) — Solution
Options
- A. ^ 1002 C_ 500
- B. ^ 1002 C_ 501
- C. ^ 1001 C_ 501
- D. ^ 1000 C_ 501
Answer
A. ^ 1002 C_ 500
Step-by-step solution
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
Related: Mathematics — Binomial Theorem · All PYQ Banks