In the expansion of {left( frac{a}{x} + bx ri | 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 ${\left( \frac{a}{x} + bx \right)^{12}}$ is given by $T_{r+1} = {^{12}C_r} \left( \frac{a}{x} \right)^{12-r}

Vedclass · Accepted Answer

$12a^{11}b$

Vedclass · Answer

(C) The general term in the expansion of ${\left( \frac{a}{x} + bx \right)^{12}}$ is given by $T_{r+1} = {^{12}C_r} \left( \frac{a}{x} \right)^{12-r} (bx)^r$. $T_{r+1} = {^{12}C_r} a^{12-r} x^{-(12-r)} b^r x^r$ $T_{r+1} = {^{12}C_r} a^{12-r} b^r x^{2r-12}$. To find the coefficient of $x^{-10}$,we set the exponent of $x$ equal to $-10$: $2r - 12 = -10$ $2r = 2$ $r = 1$. Substituting $r = 1$ into the general term: $T_{1+1} = {^{12}C_1} a^{12-1} b^1 x^{2(1)-12}$ $T_2 = 12 a^{11} b x^{-10}$. Thus,the coefficient of $x^{-10}$ is $12a^{11}b$.

In the expansion of ${\left( \frac{a}{x} + bx \right)^{12}}$,the coefficient of $x^{-10}$ is:

Similar Questions

If the term independent of $x$ in the expansion of $(\sqrt{a}x^2 + \frac{1}{2x^3})^{10}$ is $105$,then $a^2$ is equal to:

Number of rational terms in the expansion of $( \sqrt{2} + \sqrt[4]{3} )^{100}$ is:

The middle term in the expansion of $(1 + x)^{2n}$ is

If the ratio of the $7^{th}$ term from the beginning to the $7^{th}$ term from the end in the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n$ is $\frac{1}{6}$,then $n=$

The sum of all rational terms in the expansion of $(2^{\frac{1}{5}} + 5^{\frac{1}{3}})^{15}$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module