First slide

Binomial theorem for positive integral Index

Question

The term independent of $x$ in the expansion of ${(\frac{1}{2} x^{1 / 3} + x^{- 1 / 5})}^{8}$ is

Moderate

A

11
B

10
C

8
D

7

Solution

Given expansion ${(\frac{1}{2} x^{1 / 3} + x^{- 1 / 5})}^{8}$ 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 ${(x + a)}^{n})$

Here, $T_{r + 1} =^{8} C_{r} {(\frac{1}{2} x^{1 / 3})}^{8 - r} {(x^{- 1 / 5})}^{r}$

$T_{r + 1} =^{8} C_{r} {(\frac{1}{2})}^{8 - r} {(x^{1 / 3})}^{8 - r} {(x^{- 1 / 5})}^{r}$

$T_{r + 1} =^{8} C_{r} \frac{x^{\frac{8 - r}{3} - \frac{r}{5}}}{2^{8 - r}} ....................... (1)$

$x^{\frac{8 - r}{3} - \frac{r}{5}}$ compare with $x^{0}$ for finding $r$

$\Rightarrow x^{\frac{8 - r}{3} - \frac{r}{5}} = x^{0}$

which will be Independent of $x$ if $\frac{8 - r}{3} - \frac{r}{5} = 0$

$∵ r = 5$ these value in Eqn (1)

$T_{5 + 1} =^{8} C_{5} \frac{x^{\frac{8 - 5}{3} - \frac{5}{5}}}{2^{8 - 5}}$

$T_{5 + 1} =^{8} C_{5} \frac{x^{0}}{2^{3}}$

So, required term $= T_{6} =^{8} C_{5} \frac{1}{8}$

$=^{8} C_{3} \frac{1}{8} = \frac{8 \times 7 \times 6}{1 \times 2 \times 3} . \frac{1}{8} = 7$

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