(C) The general term in the expansion of $(y^2 + \frac{c}{y})^5$ is given by $T_{r+1} = ^5C_r (y^2)^{5-r} (\frac{c}{y})^r$.$T_{r+1} = ^5C_r y^{10-2r}

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(C) The general term in the expansion of $(y^2 + \frac{c}{y})^5$ is given by $T_{r+1} = ^5C_r (y^2)^{5-r} (\frac{c}{y})^r$.$T_{r+1} = ^5C_r y^{10-2r} c^r y^{-r} = ^5C_r c^r y^{10-3r}$.To find the coefficient of $y$,we set the exponent of $y$ equal to $1$:$10 - 3r = 1$$3r = 9$$r = 3$.Substituting $r = 3$ into the coefficient expression $^5C_r c^r$:Coefficient $= ^5C_3 c^3 = \frac{5 \times 4}{2 \times 1} c^3 = 10c^3$.

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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If the expansion of $(y^2 + \frac{c}{y})^5$,the coefficient of $y$ is:

A
$20c$
B
$10c$
C
$10c^3$
D
$20c^2$

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Chapter

Binomial Theorem

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

If the maximum value of the term independent of $t$ in the expansion of $\left( t^{2} x^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t} \right)^{15}$,$x \geq 0$,is $K$,then $8K$ is equal to $....$

DifficultJEE MAIN 2022

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The positive value of $a$ such that the coefficient of $x^5$ is equal to that of $x^{15}$ in the expansion of $(x^2 + \frac{a}{x^3})^{10}$ is

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The sum of the coefficients of the first $50$ terms in the binomial expansion of $(1-x)^{100}$ is equal to

DifficultJEE MAIN 2023

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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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The coefficient of $x^4$ in the expansion of $\frac{1}{(1-x)(1-2x)(1-3x)}$ is

MediumAP EAMCET 2019

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If the expansion of $(y^2 + \frac{c}{y})^5$,the coefficient of $y$ is:

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If the maximum value of the term independent of $t$ in the expansion of $\left( t^{2} x^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t} \right)^{15}$,$x \geq 0$,is $K$,then $8K$ is equal to $....$

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