If the coefficient of the 3^{text{rd}} term f | 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 $(x+y)^n$ is $T_{r+1} = ^nC_r x^{n-r} y^r$. For the expansion of $\left(ax^2 - \frac{8}{bx}\right)^9$,the $3^

Vedclass · Accepted Answer

$a^5b^5 = -2$

Vedclass · Answer

(C) The general term in the expansion of $(x+y)^n$ is $T_{r+1} = ^nC_r x^{n-r} y^r$. For the expansion of $\left(ax^2 - \frac{8}{bx}ight)^9$,the $3^{	ext{rd}}$ term from the beginning is $T_3$ $(r=2)$: $T_3 = ^9C_2 (ax^2)^{9-2} (-\frac{8}{bx})^2 = ^9C_2 a^7 x^{14} \cdot \frac{64}{b^2 x^2} = ^9C_2 \cdot \frac{64 a^7}{b^2} x^{12}$. The coefficient is $^9C_2 \cdot \frac{64 a^7}{b^2}$. For the expansion of $\left(ax - \frac{2}{bx^2}ight)^9$,the $3^{	ext{rd}}$ term from the end is the $3^{	ext{rd}}$ term from the beginning in the expansion of $\left(-\frac{2}{bx^2} + axight)^9$,which is $T_3$ $(r=2)$: $T_3 = ^9C_2 (-\frac{2}{bx^2})^{9-2} (ax)^2 = ^9C_2 (-\frac{2}{bx^2})^7 (ax)^2 = ^9C_2 \cdot (-\frac{128}{b^7 x^{14}}) \cdot a^2 x^2 = -^9C_2 \cdot \frac{128 a^2}{b^7 x^{12}}$. Wait,the question asks for the coefficient of the $3^{	ext{rd}}$ term from the end. In $(x+y)^n$,the $k^{	ext{th}}$ term from the end is the $(n-k+2)^{	ext{th}}$ term from the beginning. For $n=9, k=3$,this is the $(9-3+2) = 8^{	ext{th}}$ term from the beginning. $T_8 = ^9C_7 (ax)^{9-7} (-\frac{2}{bx^2})^7 = ^9C_7 (ax)^2 (-\frac{128}{b^7 x^{14}}) = ^9C_2 \cdot \frac{-128 a^2}{b^7 x^{12}}$. Equating the coefficients: $^9C_2 \cdot \frac{64 a^7}{b^2} = -^9C_2 \cdot \frac{128 a^2}{b^7}$. $\frac{a^7}{b^2} = -\frac{2 a^2}{b^7} \implies a^5 b^5 = -2$.

If the coefficient of the $3^{\text{rd}}$ term from the beginning in the expansion of $\left(ax^2 - \frac{8}{bx}\right)^9$ is equal to the coefficient of the $3^{\text{rd}}$ term from the end in the expansion of $\left(ax - \frac{2}{bx^2}\right)^9$,then the relation between $a$ and $b$ is:

Similar Questions

If the constant term in the expansion of $\left(\frac{\sqrt[5]{3}}{x}+\frac{2x}{\sqrt[3]{5}}\right)^{12}, x \neq 0$,is $\alpha \times 2^8 \times \sqrt[5]{3}$,then $25 \alpha$ is equal to :

If $C_j$ stands for ${}^nC_j$,then $\frac{C_1}{C_0} + \frac{2 \times C_2}{C_1} + \frac{3 \times C_3}{C_2} + \ldots + \frac{n \times C_n}{C_{n-1}} = $

The coefficient of $x^{50}$ in the expansion of $(1+x)^{101}(1-x+x^2)^{100}$ is

The total number of irrational terms in the binomial expansion of $(7^{1/5} - 3^{1/10})^{60}$ is:

The sum of the coefficients of the first three terms in the expansion of $(x - \frac{3}{x^2})^m$,where $x \neq 0$ and $m$ is a natural number,is $559$. Find the term of the expansion containing $x^3$. (in $x^3$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module