The coefficient of x^5 in the expansion of le | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The general term in the expansion of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$ is given by $T_{r+1} = { }^5 C_r \left(2 x^3\right)^{5-r} \left(-\frac

Vedclass · Accepted Answer

$\frac{80}{9}$

Vedclass · Answer

(C) The general term in the expansion of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$ is given by $T_{r+1} = { }^5 C_r \left(2 x^3\right)^{5-r} \left(-\frac{1}{3 x^2}\right)^r$.Simplifying the expression,we get $T_{r+1} = { }^5 C_r \cdot 2^{5-r} \cdot x^{3(5-r)} \cdot (-1)^r \cdot 3^{-r} \cdot x^{-2r} = { }^5 C_r \cdot \frac{2^{5-r}}{(-3)^r} \cdot x^{15-5r}$.To find the coefficient of $x^5$,we set the exponent of $x$ equal to $5$:$15 - 5r = 5$$5r = 10$$r = 2$.Substituting $r = 2$ into the expression for the coefficient:Coefficient $= { }^5 C_2 \cdot \frac{2^{5-2}}{(-3)^2} = 10 \cdot \frac{2^3}{9} = 10 \cdot \frac{8}{9} = \frac{80}{9}$.

The coefficient of $x^5$ in the expansion of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$ is

Similar Questions

The major product of the following reaction is

In which plant group are the sporophylls arranged spirally along an axis to form distinct compact structures called strobili or cones?

The correct order of ionisation energy of $C, N, O, F$ is :-

In a rectangle $ABCD$ where $BC = 2AB$,the moment of inertia will be minimum about which axis?

Vedclass Test Series

Exam Paper Generator

Online Exam Module