If the term independent of x in the expansion | 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 $(\sqrt{a}x^2 + \frac{1}{2x^3})^{10}$ is given by:$T_{r+1} = {}^{10}C_r (\sqrt{a}x^2)^{10-r} (\frac

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The general term $T_{r+1}$ in the expansion of $(\sqrt{a}x^2 + \frac{1}{2x^3})^{10}$ is given by:$T_{r+1} = {}^{10}C_r (\sqrt{a}x^2)^{10-r} (\frac{1}{2x^3})^r$$T_{r+1} = {}^{10}C_r (\sqrt{a})^{10-r} x^{20-2r} \cdot (\frac{1}{2})^r x^{-3r}$$T_{r+1} = {}^{10}C_r (\sqrt{a})^{10-r} (\frac{1}{2})^r x^{20-5r}$For the term to be independent of $x$,the exponent of $x$ must be zero:$20 - 5r = 0 \implies r = 4$Substituting $r = 4$ into the expression for the term:$T_{4+1} = {}^{10}C_4 (\sqrt{a})^{10-4} (\frac{1}{2})^4 = 105$${}^{10}C_4 = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$$210 \cdot a^3 \cdot \frac{1}{16} = 105$$a^3 \cdot \frac{210}{16} = 105$$a^3 \cdot \frac{105}{8} = 105$$a^3 = 8 \implies a = 2$Therefore,$a^2 = 2^2 = 4$.

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:

Similar Questions

The term independent of $y$ in the expansion of $(y^{-1/6} - y^{1/3})^9$ is

The coefficient of $t^{50}$ in $(1 + t^2)^{25}(1 + t^{25})(1 + t^{40})(1 + t^{45})(1 + t^{47})$ is

Find the term independent of $x$ $(x > 0, x \neq 1)$ in the expansion of $\left[\frac{(x+1)}{\left(x^{2/3}-x^{1/3}+1\right)}-\frac{(x-1)}{(x-\sqrt{x})}\right]^{10}$.

The coefficient of $x^{256}$ in the expansion of $(1-x)^{101}(x^{2}+x+1)^{100}$ is:

If coefficients of $2^{nd}$,$3^{rd}$ and $4^{th}$ terms in the binomial expansion of $(1 + x)^n$ are in $A.P.$,then $n^2 - 9n$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module