(3x−1)15=15C0(3x)0(−1)15+15C1(3x)1(−1)14 +15C2(3x)2(−1)13+.....+15C15(3x)15(−1)0 Putting x=1, we will get the sum 215=−15C0+15C1⋅3−15C2⋅32+15C3⋅33+ ...+15C15⋅315 (1+x)15=15C0+15C1x+15C2x2+...+15C15x15 Putting x=1, we will get the sum of binomial coefficient. 215=15C0+15C1+...+15C15 (1+x)16+(1−x)16=(16C0+16C1x+...(i) +(16C0−16C1x+......+16C16x16) =2(16C0+16C2x16x16) Putting x=1 216=2(16C0+16C2+16C4+...+16C16) . . . (ii) 215=16C0+16C2+...+16C16...(ii) (1+x)16−(1−x)16=(16C0+16C1x+...+16C16x16) =16C1x+...+16C16x16) =2(16C1x+16C3x3+16C5x5+...+16C15x15) Putting x=1 216=2(16C1+16C3+...+16C15) 215=16C1+16C3+...+16C15 Hence, all are correct and option (a) is the correct option.