JEE Main 2018 — Binomial Theorem Question with Solution
From: JEE Main 2018 (Online) 15th April Evening Slot
Question
The coefficien of x10 in the expansion of (1 + x)2(1 + x2)3(1 + x3)4 is equal to :
Choose an option
Show full solutionCorrect option: A
Correct answer
A52
Step-by-step explanation
(1 + x)2 = 1 + 2x + x2,
(1 + x2)3 = 1 + 3x2 + 3x4 + x6
and (1 + x3)4 = 1 + 4x3 + 6x6 + 4x9 + x12
So, the possible combination for x10 are :
x . x9, x . x6. x3, x2 . x2 . x6 , x4 . x6
Corresponding coefficients are 2 4, 2 1 4, 1 3 6,
3 6 or 8, 8, 18, 18.
Sum of the coefficient is 8 + 8 + 18 + 18 = 52
Therefore, the coefficient of x10 in the expansion of
(1 + x)2 (1 + x2)3 (1 + x3)4 is 52.
(1 + x2)3 = 1 + 3x2 + 3x4 + x6
and (1 + x3)4 = 1 + 4x3 + 6x6 + 4x9 + x12
So, the possible combination for x10 are :
x . x9, x . x6. x3, x2 . x2 . x6 , x4 . x6
Corresponding coefficients are 2 4, 2 1 4, 1 3 6,
3 6 or 8, 8, 18, 18.
Sum of the coefficient is 8 + 8 + 18 + 18 = 52
Therefore, the coefficient of x10 in the expansion of
(1 + x)2 (1 + x2)3 (1 + x3)4 is 52.
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This is a previous-year question from JEE Main 2018, covering the Binomial Theorem chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.