If the ratio of the 7^{th} term from the begi | vedclass

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(C) The general term in the expansion of $\left(2^{1/3} + 3^{-1/3}\right)^n$ is given by $T_{r+1} = { }^n C_r (2^{1/3})^{n-r} (3^{-1/3})^r = { }^n C_r

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) The general term in the expansion of $\left(2^{1/3} + 3^{-1/3}\right)^n$ is given by $T_{r+1} = { }^n C_r (2^{1/3})^{n-r} (3^{-1/3})^r = { }^n C_r 2^{(n-r)/3} 3^{-r/3}$.The $7^{th}$ term from the beginning is $T_7 = { }^n C_6 2^{(n-6)/3} 3^{-6/3} = { }^n C_6 2^{(n-6)/3} 3^{-2}$.The $7^{th}$ term from the end is the $(n-7+1+1)^{th} = (n-5)^{th}$ term from the beginning,which is $T_{n-5} = { }^n C_{n-6} 2^{(n-(n-6))/3} 3^{-(n-6)/3} = { }^n C_6 2^{6/3} 3^{-(n-6)/3} = { }^n C_6 2^2 3^{(6-n)/3}$.Given the ratio is $\frac{1}{6}$:$\frac{{ }^n C_6 2^{(n-6)/3} 3^{-2}}{{ }^n C_6 2^2 3^{(6-n)/3}} = \frac{1}{6}$$\frac{2^{(n-6)/3}}{2^2} \cdot \frac{3^{-2}}{3^{(6-n)/3}} = \frac{1}{6}$$2^{(n-6)/3 - 2} \cdot 3^{-2 - (6-n)/3} = 6^{-1}$$2^{(n-12)/3} \cdot 3^{(n-12)/3} = 6^{-1}$$(2 \cdot 3)^{(n-12)/3} = 6^{-1}$$6^{(n-12)/3} = 6^{-1}$Equating the exponents: $\frac{n-12}{3} = -1$ $\Rightarrow n-12 = -3$ $\Rightarrow n = 9$.Thus,the correct option is $C$.

If the ratio of the $7^{th}$ term from the beginning to the $7^{th}$ term from the end in the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n$ is $\frac{1}{6}$,then $n=$

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