In the expansion of 3-2-1-3-3n the ratio of 7th term from the beginning and from the end is 6667- then n-

The correct option is D 2013 giben binomial expression (3√(2) +13√(3))^n So, according to question #160; ^nC_6 (3√(2))^n 6 (13√(3))^6^nC_6 #16 ...

In the expans...

Question

In the expansion of ${(3 \sqrt{2} + \frac{1}{3 \sqrt{3}})}^{n}$ the ratio of $7^{t h}$ term from the beginning and from the end is $6^{667}$ , then $n$ =.....

667

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2001

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6

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2013

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Solution

The correct option is D $2013$
giben binomial expression ${(3 \sqrt{2} + \frac{1}{3 \sqrt{3}})}^{n}$
So, according to question
$\frac{^{n} C_{6} (3 \sqrt{2})^{n - 6} {(\frac{1}{3 \sqrt{3}})}^{6}}{^{n} C_{6} {(\frac{1}{3 \sqrt{3}})}^{n - 6} (3 \sqrt{2})^{6}}$
$\Rightarrow (3 \sqrt{2})^{n - 12} (3 \sqrt{3})^{n - 12} = (6)^{667}$
$\Rightarrow (3 \sqrt{2} 3 \sqrt{3})^{n - 12} = (6)^{667}$
$\Rightarrow (6)^{\frac{n - 12}{3}} = (6)^{667}$
$∵$ base is same $\frac{n - 12}{3} = 667 \Rightarrow n = 2013$
So, option $(d)$ is correct

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In the expansion of (3√2+13√3)n the ratio of 7th term from the beginning and from the end is 6667, then n=.....

The correct option is D 2013giben binomial expression (3√2+13√3)nSo, according to question nC6(3√2)n−6(13√3)6nC6(13√3)n−6(3√2)6⇒ (3√2)n−12(3√3)n−12=(6)667⇒ (3√2 3√3)n−12=(6)667⇒ (6)n−123=(6)667∵ base is same n−123=667⇒ n=2013So, option (d) is correct

In the expansion of ${(3 \sqrt{2} + \frac{1}{3 \sqrt{3}})}^{n}$ the ratio of $7^{t h}$ term from the beginning and from the end is $6^{667}$ , then $n$ =.....