If the coefficient of x^9 in (alpha x^3 + fra | vedclass

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

Question

(C) The general term $T_{r+1}$ in the expansion of $(ax^p + bx^q)^n$ is given by $T_{r+1} = {}^{n}C_r (ax^p)^{n-r} (bx^q)^r$.For the first expression

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) The general term $T_{r+1}$ in the expansion of $(ax^p + bx^q)^n$ is given by $T_{r+1} = {}^{n}C_r (ax^p)^{n-r} (bx^q)^r$.For the first expression $(\alpha x^3 + \frac{1}{\beta x})^{11}$,the general term is $T_{r+1} = {}^{11}C_r (\alpha x^3)^{11-r} (\beta^{-1} x^{-1})^r = {}^{11}C_r \alpha^{11-r} \beta^{-r} x^{33-3r-r} = {}^{11}C_r \alpha^{11-r} \beta^{-r} x^{33-4r}$.Setting $33-4r = 9$,we get $4r = 24$,so $r = 6$. The coefficient is ${}^{11}C_6 \alpha^5 \beta^{-6}$.For the second expression $(\alpha x - \frac{1}{\beta x^3})^{11}$,the general term is $T_{k+1} = {}^{11}C_k (\alpha x)^{11-k} (-\beta^{-1} x^{-3})^k = {}^{11}C_k \alpha^{11-k} (-1)^k \beta^{-k} x^{11-k-3k} = {}^{11}C_k \alpha^{11-k} (-1)^k \beta^{-k} x^{11-4k}$.Setting $11-4k = -9$,we get $4k = 20$,so $k = 5$. The coefficient is ${}^{11}C_5 \alpha^6 (-1)^5 \beta^{-5} = -{}^{11}C_5 \alpha^6 \beta^{-5}$.Equating the two coefficients:${}^{11}C_6 \alpha^5 \beta^{-6} = -{}^{11}C_5 \alpha^6 \beta^{-5}$.Since ${}^{11}C_6 = {}^{11}C_5$,we have $\alpha^5 \beta^{-6} = -\alpha^6 \beta^{-5}$.Dividing both sides by $\alpha^5 \beta^{-5}$,we get $\beta^{-1} = -\alpha$,which implies $\alpha \beta = -1$.Therefore,$(\alpha \beta)^2 = (-1)^2 = 1$.

If the coefficient of $x^9$ in $(\alpha x^3 + \frac{1}{\beta x})^{11}$ and the coefficient of $x^{-9}$ in $(\alpha x - \frac{1}{\beta x^3})^{11}$ are equal,then $(\alpha \beta)^2$ is equal to $.............$.

Similar Questions

If the coefficient of the $4^{th}$ term in the expansion of $(a + b)^n$ is $56$,then $n$ is

The sum of the real values of $x$ for which the middle term in the binomial expansion of ${\left( {\frac{{{x^3}}}{3} + \frac{3}{x}} \right)^8}$ equals $5670$ is

If the coefficients of $(r-5)^{th}$ and $(2r-1)^{th}$ terms in the expansion of $(1+x)^{34}$ are equal,find $r$.

If the three consecutive coefficients in the expansion of $(1 + x)^n$ are $28, 56$ and $70,$ then the value of $n$ is

Show that the middle term in the expansion of $(1+x)^{2n}$ is $\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} 2^n x^n$,where $n$ is a positive integer.

Vedclass Test Series

Exam Paper Generator

Online Exam Module