First slide

Binomial theorem for positive integral Index

Question

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

Easy

A

$\frac{5}{3}$
B

$\frac{4}{5}$
C

6
D

$\frac{1}{2}$

Solution

$The general term is, T_{r + 1} = {}^{10}C_{r} {(\frac{x}{3})}^{\frac{10 - r}{2}} \cdot \frac{3^{\frac{r}{2}}}{x^{2 r}} = {}^{10}C_{r} {(\frac{1}{3})}^{\frac{10 - r}{2}} \times 3^{r / 2} \times x^{(\frac{10 - r}{2} - 2 r)} Therefore, for independent term of x, we have, \frac{10 - r}{2} - 2 r = 0 \Rightarrow r = 2 Therefore, term independent of x = C (10, 2) {(\frac{1}{3})}^{4} \times 3 = 45 \times \frac{1 \times 3}{81} = \frac{45}{27} = \frac{5}{3}$

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