BITSAT 2018 Paper

Given, (x−

1

x

)7 and the (r+1) th term in the expansion of (x+a)n is Tr+1 =

n

‌

Cr(x)n−r
∴ (r+1) th term in the expansion of (x−

1

x

)7 =

7

‌

Cr(x)7−r(−

1

x

)r
=

7

‌

Cr(x)7−2r(−1)r
Since x3 occurs in Tr+1
∴ 7 - 2r = 3 ⇒ r = 2
thus the coefficient of r3 =

7

‌

C2(−1)2
=

7×6

2×1

= 21