If the coefficient of xn in (1+x)1011−x+x2100 is nonzero, then n cannot be of the form

We have,(1+x)1011−x+x2100=(1+x)(1+x)1−x+x2100=(1+x)1+x3100=(1+x)C0+C1x3+C2x6+⋯+C100x300,           where Cr=100Cr=(1+x)∑r=0n nCrx3r=∑r=0n nCrx3r+∑r=0n nCrx3r+1Hence, there will be no term containing 3r + 2.