Question

The value of C99 is =?

Answer 1

Option (C) `-2^99`

Explanation:

The sum of roots of given `n`th degree equation is given as

`\ ^99C_0+^99C_1+^99C_2+\dots +^99C_99=-\frac{\text{coefficient of}\ x^99}{\text{coefficient of}\ x^100}`

`2^99=-\frac{a_99}{1}\quad (\because \ \sum_{k=0}^n \ ^nC_k=2^n)`

`a_99=-2^99`