The coefficient of x^4 in {left[ {frac{x}{2}, | vedclass

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

Question

$T_{r+1}={ }^{n} C_{r} a^{n-r} b^{r}$

Applying to the above question we get $T_{r+1}=(-1)^{r+10} C_{r} x^{10-r} 2^{r-10} 3^{r} x^{-2 r}$

$=(-1)^

Vedclass · Accepted Answer

$\frac{{405}}{{256}}$

Vedclass · Answer

$T_{r+1}={ }^{n} C_{r} a^{n-r} b^{r}$

Applying to the above question we get $T_{r+1}=(-1)^{r+10} C_{r} x^{10-r} 2^{r-10} 3^{r} x^{-2 r}$

$=(-1)^{r \cdot 10} C_{r} x^{10-3 r} 2^{r-10} 3^{r} \ldots$

For the coefficient of $x^{4}$

$10-3 r=4$

$6=3 r$

$r=2$

Substituting in (i) we get $T_{3}={ }^{10} C_{2} x^{4} 2^{-8} 3^{2}$

$=\frac{10.9 .3^{2}}{2 ! 2^{8}}$

$=\frac{405}{256}$

The coefficient of $x^4$ in ${\left[ {\frac{x}{2}\,\, - \,\,\frac{3}{{{x^2}}}} \right]^{10}}$ is :

Similar Questions

The term independent of $x$ in the expansion of ${\left( {2x - \frac{3}{x}} \right)^6}$ is

If ${x^4}$ occurs in the ${r^{th}}$ term in the expansion of ${\left( {{x^4} + \frac{1}{{{x^3}}}} \right)^{15}}$, then $r = $

If the coefficients of $x^{7}$ in $\left(x^{2}+\frac{1}{b x}\right)^{11}$ and $x^{-7}$ in $\left(x-\frac{1}{b x^{2}}\right)^{11}, b \neq 0$, are equal, then the value of $b$ is equal to:

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

Prove that $\sum\limits_{r = 0}^n {{3^r}{\,^n}{C_r} = {4^n}} $