- Home
- Standard 11
- Mathematics
7.Binomial Theorem
normal
The coefficient of $x^4$ in ${\left[ {\frac{x}{2}\,\, - \,\,\frac{3}{{{x^2}}}} \right]^{10}}$ is :
A
$\frac{{405}}{{256}}$
B
$\frac{{504}}{{259}}$
C
$\frac{{450}}{{263}}$
D
$\frac{{405}}{{512}}$
Solution
$T_{r+1}={ }^{n} C_{r} a^{n-r} b^{r}$
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 $T_{3}={ }^{10} C_{2} x^{4} 2^{-8} 3^{2}$
$=\frac{10.9 .3^{2}}{2 ! 2^{8}}$
$=\frac{405}{256}$
Standard 11
Mathematics
