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BITSAT 2018 Paper

Section: Mathematics

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Question : 91 of 150

Marks: +1, -0

The coefficient of

x^3

in the expansion of

\left(x-\frac{1}{x}\right)^7

is :

14
21
28
35

Solution:

Given,

\left(x-\frac{1}{x}\right)^7

and the (r+1) th term in the expansion of

(x+a)^n

is

T_{r+1}

=

\,{}^{n}C_r (x)^{n-r}

∴ (r+1) th term in the expansion of

\left(x-\frac{1}{x}\right)^7

=

\,{}^{7}C_r (x)^{7-r} \left(-\frac{1}{x}\right)^r

=

\,{}^{7}C_r (x)^{7-2r}(-1)^r

Since

x^3

occurs in

T_{r+1}

∴ 7 - 2r = 3 ⇒ r = 2

thus the coefficient of

r^3

=

\,{}^{7}C_2 (-1)^2

=

\frac{7\times6}{2\times1}

= 21

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