In the expansion of , if the ratio of term from the beginning to the term from the end is , then the value of is [2025]

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Solution

Q.
In the expansion of ${(\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}})}^{n}, n \in N$ , if the ratio of $15^{th}$ term from the beginning to the $15^{th}$ term from the end is $\frac{1}{6}$ , then the value of ${}^{n}C_{3}$ is [2025]

1 4960

2 4060

3 1040

4 2300

Ans.
(4)

Given, ${(\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}})}^{n}, n \in N$

Now, $T_{r + 1} = {}^{n}C_{r} \cdot {(\sqrt[3]{2})}^{n - r} {(\frac{1}{\sqrt[3]{3}})}^{r}$

$∴ T_{15} = {}^{n}C_{14} \cdot {(\sqrt[3]{2})}^{n - 14} {(\frac{1}{\sqrt[3]{3}})}^{14}$

$15^{th}$ term from end = $T_{n - 13}$ from beginning

$\Rightarrow T_{n - 13} = {}^{n}C_{n - 14} \cdot {(\sqrt[3]{2})}^{14} {(\frac{1}{\sqrt[3]{3}})}^{n - 14}$

But $\frac{T_{15}}{T_{n - 13}} = \frac{{}^{n}C_{14} {(\sqrt[3]{2})}^{n - 14} {(\frac{1}{\sqrt[3]{3}})}^{14}}{{}^{n}C_{n - 14} {(\sqrt[3]{2})}^{14} {(\frac{1}{\sqrt[3]{3}})}^{n - 14}} = \frac{1}{6}$

$\Rightarrow {(\sqrt[3]{2})}^{n - 28} \times {(\sqrt[3]{3})}^{n - 28} = 6^{- 1} \Rightarrow {(\sqrt[3]{6})}^{n - 28} = 6^{- 1}$

$\Rightarrow \frac{n - 28}{3} = - 1 \Rightarrow n = 25$

So, ${}^{n}C_{3} = {}^{25}C_{3} = 2300$ .

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