The value of C99 is =?

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1 Answer

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#