First slide

Binomial theorem for positive integral Index

Question

The ratio of the coefficient of $x^{10} in {(1 - x^{2})}^{10}$ and the term independent of x $in {(x - \frac{2}{x})}^{10}$ ,is

Easy

A

1:16
B

1:32
C

1:64
D

None of these

Solution

Given, ${(1 - x^{2})}^{10} and {(x - \frac{2}{x})}^{10}$
To find Ratio of coefficient of $x^{10} in {(1 - x^{2})}^{10}$ and the term of independent of $xin {(x - \frac{2}{x})}^{10}$
Now, let $(r + 1) th$ term in ${(1 - x^{2})}^{10}$ contains $x^{10}$
Thus, coefficient of $x^{10} is -^{10} C_{5}$
Again, let term be independent of x in ${(x - \frac{2}{x})}^{10}$
$\begin{aligned} ∴ T_{r + 1} =^{10} C_{r} (x)^{10 - r} {(- \frac{2}{x})}^{r} \\ = {(- 1)^{r}}^{10} C_{r} (2)^{r} (x)^{10 - 2 r} \end{aligned}$
To get term independent of x, put 10 - 2r =0 =r r = b
Thus, the term independent of X in ${(x - \frac{2}{x})}^{10} is -^{10} C_{5} (2)^{5}$
.'. Required ratio $= \frac{-^{10} C_{5}}{-^{10} C_{5} (2)^{5}} = \frac{1}{(2)^{5}} = \frac{1}{32} = 1 : 32$

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