JEE Main 2024MathematicsBinomial TheoremHardNumerical

JEE Main 2024 — Binomial Theorem Question with Solution

JEE Main 2024 (31 Jan Shift 2)

Question

Let the coefficient of xr in the expansion of x+3n−1+x+3n−2x+2+x+3n−3x+22+....+x+2n−1 be αr. If∑r=0nαr=βn−γn,β,γ∈N, then the value of β2+γ2 equals _______.

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Show full solutionCorrect answer: 25
Correct answer
25

Step-by-step explanation

Let,

S=x+3n−1+x+3n−2x+2+x+3n−3x+22+.....+x+2n−1

Now, using the sum of G.P formula we get,

⇒S=x+3n-11-x+2x+3n1-x+2x+3

⇒S=x+3n-x+2n

Now, finding the sum of coefficients we get,

⇒∑αr=∑x+3n-∑x+2n

⇒4n−3n=βn−γn by taking x=1

So, on comparing both side we get, β=4, γ=3

Hence, β2+γ2=16+9=25

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About this question

This is a previous-year question from JEE Main 2024, 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.