(C) The general term $T_{r+1}$ in the expansion of $(2x + \frac{1}{3x})^6$ is given by $T_{r+1} = ^6C_r (2x)^{6-r} (\frac{1}{3x})^r$. For the term to

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(C) The general term $T_{r+1}$ in the expansion of $(2x + \frac{1}{3x})^6$ is given by $T_{r+1} = ^6C_r (2x)^{6-r} (\frac{1}{3x})^r$. For the term to be independent of $x$,the power of $x$ must be $0$. $x^{6-r} \cdot x^{-r} = x^{6-2r} = x^0$ $\Rightarrow 6-2r = 0$ $\Rightarrow r = 3$. Substituting $r = 3$ into the general term: $T_{3+1} = ^6C_3 (2)^3 (\frac{1}{3})^3 = 20 \times 8 \times \frac{1}{27} = \frac{160}{27}$.

Class 11 Mathematics Binomial Theorem General term, Coefficient of any power of x, Independent term, Middle term and Greatest term and Greatest coefficient

Easy

The term independent of $x$ in the expansion of ${\left( {2x + \frac{1}{{3x}}} \right)^6}$ is

A
$\frac{160}{9}$
B
$\frac{80}{9}$
C
$\frac{160}{27}$
D
$\frac{80}{3}$

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Binomial Theorem

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General term, Coefficient of any power of x, Independent term, Middle term and Greatest term and Greatest coefficient

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DifficultAP EAMCET 2013

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If the coefficients of the middle terms in the binomial expansions of $(1+\alpha x)^{26}$ and $(1-\alpha x)^{28}$,where $\alpha \neq 0$,are equal,then the value of $\alpha$ is:

DifficultJEE MAIN 2026

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MediumTS EAMCET 2024

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EasyTS EAMCET 2023

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The term independent of $x$ in the expansion of ${\left( {2x + \frac{1}{{3x}}} \right)^6}$ is

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