First slide

Binomial theorem for positive integral Index

Question

The term void of $x$ in the expansion of ${(\sqrt{\frac{x}{3}} + \frac{\sqrt{3}}{x^{2}})}^{10}$ is

Moderate

A

$\frac{5}{9}$
B

$\frac{25}{9}$
C

$\frac{5}{3}$
D

None of these

Solution

        Given expansion is ${(\sqrt{\frac{x}{3}} + \frac{\sqrt{3}}{x^{2}})}^{10}$ then we have to find independent term
General term in the expansion $T_{r + 1} =^{n} C_{r} x^{n - r} a^{r}$

$T_{r + 1} =^{10} C_{r} {(\sqrt{\frac{x}{3}})}^{10 - r} {(\frac{\sqrt{3}}{x^{2}})}^{r}$ (simplication of $x$ terms and constant)

          $=^{10} C_{r} x^{\frac{10 - r}{2} - 2 r} 3^{r - 5}$ ( these $x$ term compare with $x^{0}$ )

This term is independent of $x$ if $\begin{matrix} x^{\frac{10 - r}{2} - 2 r} = x^{0} \end{matrix}$ powers are equal then we get

i.e. if $r = 2$ hence

$\begin{matrix} T_{2 + 1} =^{10} C_{2} {(\frac{\sqrt{x}}{3})}^{10 - 2} {(\frac{\sqrt{3}}{x^{2}})}^{2} \\ \Rightarrow T_{3} =^{10} C_{2} {(\frac{\sqrt{x}}{3})}^{8} {(\frac{\sqrt{3}}{x^{2}})}^{2} \end{matrix}$

   $=^{10} C_{2} 3^{- 3} = \frac{10 \times 9}{2} \times \frac{1}{27} = \frac{5}{3} (∴^{n} C_{r} = \frac{n!}{(n - r)! r!})$

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