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JEE Main 202524 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Binomial Theorem

For some n ≠ 10 , let the coefficients of the 5 th, 6 th and 7 th...

For some n≠10n \neq 10, let the coefficients of the 5 th, 6 th and 7 th terms in the binomial expansion of (1+x)n+4(1+\mathrm{x})^{\mathrm{n}+4} be in A.P. Then the largest coefficient in the expansion of (1+x)n+4(1+\mathrm{x})^{\mathrm{n}+4} is:

Options

  1. A2020
  2. B1010
  3. C3535
  4. D7070

Correct answer

C. 3535

Step-by-step solution

(1+x)n+4n+4C4,n+4C5,n+4C6,→ A.P. ⇒2×n+4C5=n+4C4+n+4C6⇒4×n+4C5=(n+4C4+n+4C5)+(n+4C5+n+4C6)⇒4×n+4C5=n+5C5+n+5C6\begin{aligned} & (1+x)^{n+4} \\ & { }^{n+4} C_4,{ }^{n+4} C_5,{ }^{n+4} C_6, \rightarrow \text{ A.P. } \\ & \Rightarrow 2 \times { }^{n+4} C_5={ }^{n+4} C_4+{ }^{n+4} C_6 \\ & \Rightarrow 4 \times{ }^{n+4} C_5=\left({ }^{n+4} C_4+{ }^{n+4} C_5\right)+\left({ }^{n+4} C_5+{ }^{n+4} C_6\right) \\ & \Rightarrow 4 \times{ }^{n+4} C_5={ }^{n+5} C_5+{ }^{n+5} C_6\end{aligned}⇒4×(n+4)!5!⋅(n−1)!=(n+6)!6!⋅n!⇒4=(n+6)(n+5)6n⇒n2+11n+30=24n⇒n2−13n+30=0⇒n=3,10( rejected )∵n≠10\begin{aligned} & \Rightarrow 4 \times \frac{(\mathrm{n}+4)!}{5!\cdot(\mathrm{n}-1)!}=\frac{(\mathrm{n}+6)!}{6!\cdot \mathrm{n}!} \\ & \Rightarrow 4=\frac{(\mathrm{n}+6)(\mathrm{n}+5)}{6 \mathrm{n}} \\ & \Rightarrow \mathrm{n}^2+11 \mathrm{n}+30=24 \mathrm{n} \\ & \Rightarrow \mathrm{n}^2-13 \mathrm{n}+30=0 \\ & \Rightarrow \mathrm{n}=3,10(\text{ rejected }) \\ & \because \mathrm{n} \neq 10\end{aligned} \thereforeLargestbinomialcoefficientinexpansionofLargest binomial coefficient in expansion of(1+x)7(∵n+4=7)\begin{aligned} & (1+\mathrm{x})^7 \\ & (\because \mathrm{n}+4=7) \end{aligned}iscoeff.ofmiddletermis coeff. of middle term\Rightarrow{ }^7 \mathrm{C}_4={ }^7 \mathrm{C}_3=35$
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