UPSC NDA Math Model Paper 10

Given: (1−x)k and the third term of the expansion is (−1∕4)x2
Expansion of (1−x)k=[‌kC0−‌kC1x+‌kC2x2+.....]
So the third term =‌kC2x2=(−1∕4)x2
⇒

k!

2!(k−2)!

=

1

4

⇒

k×(k−1)×(k−2)!

2(k−2)!

=

1

4

⇒

k×(k−1)

2

=

1

4

⇒k2−k=

1

2

⇒2k2−2k−1=0 .....(Multiply by 2 on both the sides)
⇒k=

2±√12

4

=

1±√3

2

⇒k=

1±√3

2

Hence, option (1) is correct.