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7.Binomial Theorem
medium
$\left(2 x^3-\frac{1}{3 x^2}\right)^5$ ના વિસ્તરણમાં $x^5$ નો સહગુણક $........$ હશે.
A
$8$
B
$9$
C
$\frac{80}{9}$
D
$\frac{26}{3}$
(JEE MAIN-2023)
Solution
$\left(2 x^3-\frac{1}{3 x^2}\right)^5$
$T_{r+1}={ }^5 C_r\left(2 x^3\right)^{5-r}\left(\frac{-1}{3 x^2}\right)^r={ }^5 C_r \frac{(2)^{5-r}}{(-3)^r}(x)^{15-5 r}$
$\therefore 15-5 r =5$
$\therefore r =2$
$T_3=10\left(\frac{8}{9}\right) x^5$
So, coefficient is $\frac{80}{9}$
Standard 11
Mathematics
