First slide

Binomial theorem for positive integral Index

Question

Coefficient of the term independent of $x$ in the expansion of ${(\frac{x + 1}{x^{2 / 3} - x^{1 / 3} + 1} - \frac{x - 1}{x - x^{1 / 2}})}^{10}$ is

Moderate

A

$210$
B

$105$
C

$70$
D

$35$

Solution

We have
$\begin{array}{l} x + 1 = {(x^{1 / 3})}^{3} + 1 = (x^{1 / 3} + 1) (x^{2 / 3} - x^{1 / 3} + 1), \\ x - 1 = (\sqrt{x} - 1) (\sqrt{x} + 1) \end{array}$
and $x - x^{1 / 2} = \sqrt{x} (\sqrt{x} - 1),$
Thus, $\frac{x + 1}{x^{2 / 3} - x^{1 / 3} + 1} - \frac{x - 1}{x - x^{1 / 2}}$
$= x^{1 / 3} + 1 - (1 + x^{- 1 / 2}) = x^{1 / 3} - x^{- 1 / 2}$
Now, the $(r + 1) t h$ term in the expansion of ${(x^{1 / 3} - x^{- 1 / 2})}^{10}$ is
$\begin{aligned} T_{r + 1} =^{10} C_{r} {(x^{1 / 3})}^{10 - r} {(x^{- 1 / 2})}^{r} (- 1)^{r} \\ =^{10} C_{r} x^{10 \sqrt{3} - 5 r / 6} (- 1)^{r} \end{aligned}$
For this term to be independent of $x$ ,
$\frac{10}{3} - \frac{5 r}{6} = 0 \Rightarrow r = \frac{10}{3} \times \frac{6}{5} = 4$
Thus, coefficient of the term independent of $x$ in the given expression is
$^{10} C_{4} = \frac{10!}{4! 6!} = 210 .$

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