the coefficirent of x^99 in (x+1) (x+3) (x+5) ......(x+199) is?????

The answer for this question is  1+3+5+7+ ... +197+199

This is so because the coefficient of the term with exponent one less than the degree, in an expansion is always equal to the sum of the constant terms in the linear factors.

Here the constant terms in the linear factors are 1,3,5,7,9 ... ,199

Therefore the coefficient is 1+3+5+...199 = (1+199)100/2 (Sum of a.p)= 10000 ---(Ans)