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  • If the seventh term from the beginning is equal to the seventh term from the end in the expansion of <img alt="\left[2^{1 / 3}+\frac{1}{3^{1 / 3}}\right]^n," src="https://entrancecorner.oncode

If the seventh term from the beginning is equal to the seventh term from the end in the expansion of \left[2^{1 / 3}+\frac{1}{3^{1 / 3}}\right]^n, then n=

Option: 1

12


Option: 2

13


Option: 3

14


Option: 4

10


Answers (1)

best_answer

7th term from the end in the expansion of \left(2^{1 / 3}+3^{-1 / 3}\right)^n is

(n+1-6) \text { th }=(n-5) \text { th term. } 

We have T_7=T_{6+1}={ }^n C_6\left(2^{1 / 3}\right)^6\left(3^{-1 / 3}\right)^{n-6}={ }^n C_6 2^2 \cdot 3^{2-(n / 3)}

and T_{n-5}=T_{(n-6)+1}={ }^n C_{n-6}\left(2^{1 / 3}\right)^{n-6}\left(3^{-1 / 3}\right)^6={ }^n C_6 2^{(n / 3)-2} 3^{-2}

Since T_7=T_{n-5}, we get

{ }^n C_6 2^2 3^{-(n / 3)+2}={ }^n C_6 2^{(n / 3)-2} 3^{-2}

\Rightarrow 2^2 3^{-(n / 3)+2}=2^{(n / 3)-2} 3^{-2} \Rightarrow 2^{(n / 3)-4} 3^{(n / 3)-4}=1 \Rightarrow 6^{(n / 3)-4}=1

But this is possible only when (n / 3)-4=0 \text { or } n=12

Posted by

himanshu.meshram

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