The correct option is A k1=k2
WehaveAk=∣∣∣c2k−100c22∣∣∣An=[c2000c21]+[c2100c22]+.....+[c2n−100c2n]=[c20+c21+...+c2n−11pt00c21+c22+...+c2n1pt]=[2ncn−1002ncn−1](1+x)n=nc0+nc1x+nc2x2+......+ncnxn(x+1)n=nc0xn+nc1xn−1+nc2xx−2+.......+ncn(1+x)2n=[nc0+nc1x+....+ncnxn][nc0xn+nc1xn−1+....+ncn]coefficientofxn:2ncn=nc20+nc20+....+nc2nK1=K2=2ncn−1Hence,optionAisthecorrectanswer.