(1+x+x2+⋯+x50)3=(1+x+x2+⋯+x50)⋅(1+x+x2+⋯+x50)⋅(1+x+x2+⋯+x50). The coefficient of x50 is just the number of xaxbxc=xa+b+c with a+b+c=50, a,b,c∈{0,1,…,50}. If a=0 the equation b+c=50 has 51 solutions, namely (0,50), (1,49), …, (50,0). Also, if a=1, the equation b+c=49 has 50 solutions and so on. The number of all solutions is 51+50+⋯+1=1326. Hence c50=1326. The answer is (C).