Mathematics · Binomial Theorem

JEE Main 2024 — 9 April, Shift 1 — Question 6

The coefficient of x70x^{70} in x2(1+x)98+x3(1+x)97+x^{2}(1+x)^{98}+x^{3}(1+x)^{97}+ x4(1+x)96+\mathrm{x}^{4}(1+\mathrm{x})^{96}+.\qquad. +x54(1+x)46+\mathrm{x}^{54}(1+\mathrm{x})^{46} is

99Cp−46Cq{ }^{99} \mathrm{C}_{\mathrm{p}}-{ }^{46} \mathrm{C}_{\mathrm{q}}. Then a possible value to p+q\mathrm{p}+\mathrm{q} is :

  1. Option A:

    55

  2. Option B:

    61

  3. Option C:

    63

  4. Option D:

    83

    Correct

Answer: D

Step-by-step solution

x2(1+x)98+x3(1+x97)+x4(1+x)96+…….\quad \mathrm{x}^{2}(1+\mathrm{x})^{98}+\mathrm{x}^{3}\left(1+\mathrm{x}^{97}\right)+\mathrm{x}^{4}(1+\mathrm{x})^{96}+\ldots \ldots ..x54(1+x)46x^{54}(1+x)^{46}

Coeff. of x70:98C68+97C67+96C66+\mathrm{x}^{70}:{ }^{98} \mathrm{C}_{68}+{ }^{97} \mathrm{C}_{67}+{ }^{96} \mathrm{C}_{66}+ \qquad +47C17+46C16+{ }^{47} \mathrm{C}_{17}+{ }^{46} \mathrm{C}_{16} =46C30+47C30+………..98C30={ }^{46} \mathrm{C}_{30}+{ }^{47} \mathrm{C}_{30}+\ldots \ldots \ldots . .{ }^{98} \mathrm{C}_{30}

=(46C31+46C30)+47C30+………..98C30−46C31=\left({ }^{46} \mathrm{C}_{31}+{ }^{46} \mathrm{C}_{30}\right)+{ }^{47} \mathrm{C}_{30}+\ldots \ldots \ldots . .{ }^{98} \mathrm{C}_{30}-{ }^{46} \mathrm{C}_{31}

=47C31+47C30+………..98C30−46C31={ }^{47} \mathrm{C}_{31}+{ }^{47} \mathrm{C}_{30}+\ldots \ldots \ldots . .{ }^{98} \mathrm{C}_{30}-{ }^{46} \mathrm{C}_{31}

=99C31−46C31=99Cp−46Cq={ }^{99} \mathrm{C}_{31}-{ }^{46} \mathrm{C}_{31}={ }^{99} \mathrm{C}_{\mathrm{p}}-{ }^{46} \mathrm{C}_{\mathrm{q}}

Possible values of (p+q)(\mathrm{p}+\mathrm{q}) are 62,83,99,4662,83,99,46

⇒p+q=83\Rightarrow \mathrm{p}+\mathrm{q}=83

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Binomial Theorem
Topic
Series involving sum & product of Binomial Coefficients