First slide

Binomial theorem for positive integral Index

Question

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

Moderate

A

$^{10} C_{1}$
B

$\frac{5}{12}$
C

$1$
D

None

Solution

Given expansion ${(\frac{\sqrt{x}}{3} + \sqrt{\frac{3}{2 x^{2}}})}^{10}$ is
We have general term in the expansion ${(x + a)}^{n}$

$(∴ T_{r + 1} =^{n} C_{r} x^{n - r} {(a)}^{r}$ be the expansion of ${(x + a)}^{n})$

General term   $T_{r + 1} =^{10} C_{r} {(\sqrt{\frac{x}{3}})}^{10 - r} {(\sqrt{\frac{3}{2 x^{2}}})}^{r}$

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

   $T_{r + 1} =^{10} C_{r} x^{\frac{10 - r}{2} - r} \frac{{(\sqrt{3})}^{r}}{{(\sqrt{3})}^{10 - r} 2^{r / 2}}$

$x^{\frac{10 - r}{2} - r}$ compare with $x^{0}$ because of independent term

$\Rightarrow x^{\frac{10 - r}{2} - r} = x^{0}$

This will not contain x if $\frac{10 - r}{2} - r = 0$

   $\Rightarrow 10 - r - 2 r = 0$

    $∵ r = \frac{10}{3}$

Since $r$ ought to be a whole number, therefore, there is no term void of $x$ in the given power of given binomial.

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