(C) Given expression is $(1 + 3x + 3x^2 + x^3)^6$.We know that $(1 + x)^3 = 1 + 3x + 3x^2 + x^3$.So,the expression becomes $((1 + x)^3)^6 = (1 + x)^{1

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(C) Given expression is $(1 + 3x + 3x^2 + x^3)^6$.We know that $(1 + x)^3 = 1 + 3x + 3x^2 + x^3$.So,the expression becomes $((1 + x)^3)^6 = (1 + x)^{18}$.For a binomial expansion $(1 + x)^n$,if $n$ is even,the middle term is given by the $(\frac{n}{2} + 1)^{th}$ term.Here,$n = 18$,which is even.Therefore,the middle term is $(\frac{18}{2} + 1)^{th} = (9 + 1)^{th} = 10^{th}$ term.

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

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The middle term in the expansion of $(1 + 3x + 3x^2 + x^3)^6$ is

A
$4^{th}$
B
$3^{rd}$
C
$10^{th}$
D
None of these

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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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DifficultJEE MAIN 2022

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Find the term independent of $x$ $(x > 0, x \neq 1)$ in the expansion of $\left[\frac{(x+1)}{\left(x^{2/3}-x^{1/3}+1\right)}-\frac{(x-1)}{(x-\sqrt{x})}\right]^{10}$.

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