If 1+x+x2+x3n=a0+a1x+a2x2+a3x3+…+a3nx3n, then the value of  a0+a4+a8+a12+… is

we have  a0+a1x+a2x2+a3x3+…+a3nx3n=1+x+x2+x3n⇒a0+a2x2+a4x4+…+a1x+a3x3+…=1+x+x2+x3nPutting x = 1, -1, i and -i respectively, we geta0+a2+a4+…+a1+a3+a5+….=4na0+a2+a4+…−a1+a3+a5+….=0a0−a2+a4−a6+…+ia1−a3+a5+….=0a0−a2+a4−a6+…−ia1−a3+a5+….=0Adding (i) and (ii), we get 2a0+a2+a4+…=4nAdding (iii) and (iv), we get2a0−a2+a4−a6+…..=0 Adding (v) and (vi), we get 4a0+a4+a8+….=4n⇒a0+a4+a8+….=4n−1