Constant term in the expansion of

Question:

Constant term in the expansion of $\left(x-\frac{1}{x}\right)^{10}$ is

(a) 152

(b) −152

(c) −252

(d) 252

Solution:

(c) −252

Suppose (r + 1)th term is the constant term in the given expansion.

Then, we have:

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

$={ }^{10} C_{r}(-1)^{r} x^{10-r-r}$

For this term to be constant, we must have:

$10-2 r=0$

$\Rightarrow r=5$

$\therefore$ Required term $=-{ }^{10} C_{5}=-252$

 

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now