aj×j (1−x+x3)n=a0+a1x+a2x2+...+a3nx3n Putting x=1, (1−1+1)n‌‌=a0+a1+a2+...+a3n 1‌‌=a0+a1+a2+...+a3n.....(i) Putting x=−1, (1+1−1)n=a0−a1+a2−a3+...(−1)3nan 1=a0−a1+a2−a3+...(−1)3nan....(ii) Adding Eqs. (i) and (ii), we get 2=2(a0+a2+a4+a6...) a0+a2+a4+...=1 On subtracting Eq. (ii) from Eq. (i), we get 0=2(a1+a3+a5+...) a1+a3+a5+...=0 ‌‌ Now,