First slide

Binomial theorem for positive integral Index

Question

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

Easy

A

$\frac{9}{4}$
B

$\frac{3}{4}$
C

$\frac{5}{4}$
D

$\frac{7}{4}$

Solution

Given expansion ${(\sqrt{\frac{x}{3}} + \frac{3}{2 x^{2}})}^{10}$ 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 ${(x + a)}^{n})$

${(r + 1)}^{t h}$ term:

$T_{r + 1} =^{n} C_{r} x^{n - r} {(a)}^{r}$

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

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

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

$T_{r + 1} =^{10} C_{r} \frac{3^{r}}{3^{\frac{(10 - r)}{2}} 2^{r}} x^{\frac{(10 - r - 4 r)}{2}} .................... (1)$

$x^{\frac{(10 - r - 4 r)}{2}}$ compare with $x^{0}$ for finding $r$

$\Rightarrow x^{\frac{(10 - r - 4 r)}{2}} = x^{0}$

For term independent of $x, 10 - 5 r = 0$

$∵ r = 2$ substitute in Eqn (1)

$∴$ The term independent of $x =^{10} C_{2} \frac{3^{2}}{3^{4} {.2}^{2}} = \frac{45}{9 \times 4} = \frac{5}{4}$ $\begin{matrix} (∴^{n} C_{r} = \frac{n!}{(n - r)! r!}) \end{matrix}$

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