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JEE Main 20254 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2025 · 4 April · Morning Shift · Mathematics · Binomial Theorem

In the expansion of ( [3] 2 + 1 [3] 3 )^n, n ∈ N , if the ratio of...

In the expansion of (23+133)n,n∈N\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n, \mathrm{n} \in \mathrm{N}, if the ratio of 15at 15^{\text{at }} term from the beginning to the 15th 15^{\text{th }} term from the end is 16\frac{1}{6}, then the value of nC3{ }^n C_3 is:

Options

  1. A4060
  2. B1040
  3. C2300
  4. D4960

Correct answer

C. 2300

Step-by-step solution

Tr+1=nCr(21/3)n−r(131/3)rr=14T15=nC14(21/3)n−14(131/3)14\begin{aligned} & \mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\left(2^{1 / 3}\right)^{\mathrm{n}-\mathrm{r}}\left(\frac{1}{3^{1 / 3}}\right)^{\mathrm{r}} \\ & \mathrm{r}=14 \\ & \mathrm{ T}_{15}={ }^{\mathrm{n}} \mathrm{C}_{14}\left(2^{1 / 3}\right)^{\mathrm{n}-14}\left(\frac{1}{3^{1 / 3}}\right)^{14} \end{aligned}\mathrm{T}_{15}^{\prime}=15^{\mathrm{h}}termfromlastisterm from last is(\mathrm{n}-13)^{\mathrm{h}}termfrombeginning.term from beginning.T15′=nCn−14(21/3)14(131/3)n−14⇒T15T15′=nC14(21/3)n−14(131/3)14nCn−14(21/3)14(131/3)n−14=16\begin{aligned} & \mathrm{T}_{15}^{\prime}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-14}\left(2^{1 / 3}\right)^{14}\left(\frac{1}{3^{1 / 3}}\right)^{\mathrm{n}-14} \\ & \Rightarrow \frac{\mathrm{ T}_{15}}{\mathrm{ T}_{15}^{\prime}}=\frac{{ }^{\mathrm{n}} \mathrm{C}_{14}\left(2^{1 / 3}\right)^{\mathrm{n}-14}\left(\frac{1}{3^{1 / 3}}\right)^{14}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-14}\left(2^{1 / 3}\right)^{14}\left(\frac{1}{3^{1 / 3}}\right)^{\mathrm{n}-14}}=\frac{1}{6}\end{aligned}$\begin{aligned} & =\left(2^{1 / 3}\right)^{n-28}\left(3^{1 / 3}\right)^{n-28}=\frac{1}{6} \\ & =6^{\frac{n-28}{3}}=6^{-1}\end{aligned} \text{ So, }{ }^n \mathrm{C}_3={ }^{25} \mathrm{C}_3=2300$
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