(A) The given expression is ${\left( \frac{1 + x}{1 - x} \right)^2} = (1 + x)^2 (1 - x)^{-2}$.Expanding $(1 + x)^2 = 1 + 2x + x^2$ and $(1 - x)^{-2} =

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(A) The given expression is ${\left( \frac{1 + x}{1 - x} \right)^2} = (1 + x)^2 (1 - x)^{-2}$.Expanding $(1 + x)^2 = 1 + 2x + x^2$ and $(1 - x)^{-2} = \sum_{k=0}^{\infty} (k+1)x^k = 1 + 2x + 3x^2 + \dots + (n-1)x^{n-2} + nx^{n-1} + (n+1)x^n + \dots$.Multiplying these,the coefficient of $x^n$ is obtained by:$1 \cdot (n+1) + 2 \cdot n + 1 \cdot (n-1) = n + 1 + 2n + n - 1 = 4n$.Thus,the coefficient of $x^n$ is $4n$.

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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In the expansion of ${\left( \frac{1 + x}{1 - x} \right)^2}$,the coefficient of ${x^n}$ is:

A
$4n$
B
$4n - 3$
C
$4n + 1$
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

The value of $x$,for which the $6^{th}$ term in the expansion of ${\left\{ {{2^{{{\log }_2}\sqrt {({9^{x - 1}} + 7)} }} + \frac{1}{{{2^{(1/5){{\log }_2}({3^{x - 1}} + 1)}}}}} \right\}^7}$ is $84$,is equal to

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MediumJEE MAIN 2020

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MediumJEE MAIN 2026

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The ratio of the $5^{th}$ term from the beginning to the $5^{th}$ term from the end in the binomial expansion of $\left( 2^{1/3} + \frac{1}{2(3)^{1/3}} \right)^{10}$ is

DifficultJEE MAIN 2019

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In the expansion of ${\left( \frac{1 + x}{1 - x} \right)^2}$,the coefficient of ${x^n}$ is:

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The value of $x$,for which the $6^{th}$ term in the expansion of ${\left\{ {{2^{{{\log }_2}\sqrt {({9^{x - 1}} + 7)} }} + \frac{1}{{{2^{(1/5){{\log }_2}({3^{x - 1}} + 1)}}}}} \right\}^7}$ is $84$,is equal to

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