The absolute difference of the coefficients o | vedclass

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

Question

(A) The general term $T_{r+1}$ in the expansion of $\left(2x^2+\frac{1}{2x}\right)^{11}$ is given by:$T_{r+1} = {}^{11}C_r (2x^2)^{11-r} \left(\frac{1

Vedclass · Accepted Answer

$12^3-12$

Vedclass · Answer

(A) The general term $T_{r+1}$ in the expansion of $\left(2x^2+\frac{1}{2x}\right)^{11}$ is given by:$T_{r+1} = {}^{11}C_r (2x^2)^{11-r} \left(\frac{1}{2x}\right)^r$$= {}^{11}C_r \cdot 2^{11-r} \cdot x^{22-2r} \cdot 2^{-r} \cdot x^{-r}$$= {}^{11}C_r \cdot 2^{11-2r} \cdot x^{22-3r}$For the coefficient of $x^{10}$,set $22-3r = 10$,which gives $3r = 12$,so $r = 4$.Coefficient of $x^{10} = {}^{11}C_4 \cdot 2^{11-8} = {}^{11}C_4 \cdot 2^3 = 330 \cdot 8 = 2640$.For the coefficient of $x^7$,set $22-3r = 7$,which gives $3r = 15$,so $r = 5$.Coefficient of $x^7 = {}^{11}C_5 \cdot 2^{11-10} = {}^{11}C_5 \cdot 2^1 = 462 \cdot 2 = 924$.The absolute difference is $|2640 - 924| = 1716$.Checking the options:$12^3 - 12 = 1728 - 12 = 1716$.Thus,the correct option is $A$.

The absolute difference of the coefficients of $x^{10}$ and $x^7$ in the expansion of $\left(2x^2+\frac{1}{2x}\right)^{11}$ is equal to

Similar Questions

The sum of the rational terms in the expansion of $(\sqrt{2}+\sqrt[5]{3})^{10}$ is

If $C_0, C_1, C_2, \ldots, C_8$ are the binomial coefficients in the expansion of $(1+x)^8$,then $\sum_{r=1}^8 r^3 \frac{C_r}{C_{r-1}} =$

The two middle terms in the expansion of ${\left( {x - \frac{1}{x}} \right)^{11}}$ are:

If the constant term in the expansion of $(1+2x-3x^3)(\frac{3}{2}x^2-\frac{1}{3x})^9$ is $p$,then $108p$ is equal to:

The coefficient of $x^5$ in the expansion of $(1+x^2)^5(1+x)^4$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module