(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

The coefficients of two consecutive terms in the expansion of $(1 + x)^n$ will be equal,if

Easy

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The $16^{th}$ term in the expansion of $(\sqrt{x} - \sqrt{y})^{17}$ is

Easy

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The term independent of $x$ in the expansion of $\left( \frac{1}{60} - \frac{x^8}{81} \right) \left( 2x^2 - \frac{3}{x^2} \right)^6$ is equal to

DifficultJEE MAIN 2019

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In the expansion of $(1+x)^n$,the coefficients of the $p^{th}$ and $(p+1)^{th}$ terms are respectively $p$ and $q$. Then $p+q$ is equal to:

MediumTS EAMCET 2002

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If the numerically greatest term in the expansion of $(2-3x)^9$ when $x=1$ is $P_1^\alpha P_2^\beta P_3^\gamma P_4^\delta$ (where $P_1 < P_2 < P_3 < P_4$ are the first four prime numbers),then $\alpha+\beta+\gamma+\delta=$

MediumAP 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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