(1−x+x3)n=j=0∑3najxj
⇒(1−x+x3)n=a0+a1x+a2x2……+a3nx3n...(i)
Then,
j=0∑[23n]a2j=a0+a2+a4……
j=0∑[23n−1]a2j+1=a1+a3+a5……
Put x=1 in (i), then we get
1=a0+a1+a2+a3………+a3n ...(A)
Put x=−1 in (i), then we get
1=a0−a1+a2−a3………+(−1)3na3n ...(B)
Solving (A) and (B)
a0+a2+a4…..=1
a1+a3+a5……=0
Hence,
j=0∑[23n]a2j+4j=0∑[23n−1]a2j+1=1