Show that the term independent of x in the expansion of
Question:

Show that the term independent of x in the expansion of $\left(x-\frac{1}{x}\right)^{10}$ is -252.

 

 

Solution:

To show: the term independent of x in the expansion of $\left(\mathrm{X}-\frac{1}{\mathrm{x}}\right)^{10}$ is -252.

Formula Used:

General term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is is given by,

$T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where

${ }^{n} C_{r}=\frac{n !}{r !(n-r) !}$

Now, finding the general term of the expression, $\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)^{10}$, we get

$T_{r+1}={ }^{10} C_{r} \times x^{10-r} \times\left(\frac{-1}{x}\right)^{r}$

For finding the term which is independent of x,

$10-2 r=5$

$r=5$

Thus, the term which would be independent of $x$ is $T_{6}$

$T_{6}={ }^{10} C_{5} \times x^{10-5} \times\left(\frac{-1}{x}\right)^{5}$

$T_{6}={ }^{10} C_{5} \times x^{10-5} \times\left(\frac{-1}{x}\right)^{5}$

$T_{6}=-{ }^{10} C_{5}$

$T_{6}=-\frac{10 !}{5 !(10-5) !}$

$T_{6}=-\frac{10 !}{5 ! \times 5 !}$

$T_{6}=-\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 \times 4 \times 3 \times 2}$

$T_{6}=-\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 \times 4 \times 3 \times 2}$

$T_{6}=252$

Thus, the term independent of $x$ in the expansion of $\left(x-\frac{1}{x}\right)^{10}$ is $-252$.

 

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