Verified Solution • 2025–26 Session moderate

QJEEM24Ma6fe1: JEE Main 2024 Mathematics - Class 11 JEEM24Ma6fe1 Solution: The coefficient of in is . Then a possible value of is :

The coefficient of x70x^{70} in x2(1+x)98+x3(1+x)97+x4(1+x)96+…+x54(1+x)46x^2(1+x)^{98}+x^3(1+x)^{97}+x^4(1+x)^{96}+\ldots+x^{54}(1+x)^{46} is 99Cp−46Cq{ }^{99} \mathrm{C}_{\mathrm{p}}-{ }^{46} \mathrm{C}_{\mathrm{q}} . Then a possible value of p+q\mathrm{p}+\mathrm{q} is :

Multiple Choice Options

  • A 61
  • B 83
  • C 55
  • D 68

Step-by-Step Verified Solution

x2(1+x)98+x3(1+x)97+…+x54(1+x)46x^2(1+x)^{98}+x^3(1+x)^{97}+\ldots+x^{54}(1+x)^{46}

It is a G.P. with first term =x2(1+x)98=x^2(1+x)^{98}

and common ratio =x1+x=\frac{x}{1+x}

sum of these term =x2(1+x)98((x1+x)53−1x1+x−1)=x^2(1+x)^{98}\left(\frac{\left(\frac{x}{1+x}\right)^{53}-1}{\frac{x}{1+x}-1}\right)

=x2(1+x)98((1+x)−x53(1+x)−52)=x^2(1+x)^{98}\left((1+x)-x^{53}(1+x)^{-52}\right)

JEE Main 2024 (Online) 9th April Morning Shift Mathematics - Binomial Theorem Question 20 English Explanation

amp;=99C68−46C15amp;⇒p=68,q=15amp;⇒p+q=83\begin{aligned}& ={ }^{99} \mathrm{C}_{68}-{ }^{46} \mathrm{C}_{15} \\& \Rightarrow p=68, q=15 \\& \Rightarrow p+q=83\end{aligned}

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