The sum of the coefficients of all the integral powers of x in (1+3√x)100 is
299 [2100 - 1]
299 [2100 + 1]
4100 + 2100
None of these
100C0 + 100C232 + 100C434 .... + 100C1003100 = 12[(3+1)100+(3−1)100]=12[4100+2100]