7.Binomial Theorem
easy

${\left( {{y^2} + \frac{c}{y}} \right)^5}$ के विस्तार में $y$ का गुणांक होगा

A

$20c$

B

$10c$

C

$10{c^3}$

D

$20{c^2}$

Solution

$2(5 -r) + (-1) r = 1 $

$\Rightarrow  10 –  2r -r = 1$

$ \Rightarrow \,r = 3$

अत: $y$ का गुणांक $^5{C_3}{c^3} = \frac{{5 \times 4}}{{2 \times 1}}{c^3} = 10{c^3}$.

Standard 11
Mathematics

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