Binomial Theorem · Question 11

JEE Main 2026 (Online) 28th January Evening ShiftMCQMedium📈 Topic in 52% papers+4−1

The sum of the coefficients of x499 and x500 in (1 + x)1000 + x(1 + x)999 + x2(1 + x)998 + + x1000 is :

1002C501
1001C501
1000C501
1002C500

Answer

(4) 1002C500

Solution

S = k=01000 xk(1 + x)1000-k is a G.P. with first term (1 + x)1000 and ratio x1 + x

S = (1 + x)1000 1 - ( x1 + x )10011 - x1 + x

= (1 + x)1001 - x1001

Coefficient of x499 = 1001C499; coefficient of x500 = 1001C500

Sum = 1001C499 + 1001C500 = 1002C500 ( nCr-1 + nCr = n+1Cr)

Answer: (4)

JEE Main 2026 (Online) 28th January Evening Shift

Keep practising — Binomial Theorem

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2100101
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JEE Main 2026 (Online) 22nd January Morning Shift

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JEE Main 2026 (Online) 22nd January Evening Shift

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