We have, (1−x+x2)n = a0+a1x+a2x2 + ... + a2nx2n ... (i) Put x = 1 in Eq. (i), we get (1−1+1)n = a0+a1+a2 + ... + a2n ⇒ a0+a1+a2 ... a2n = 1 ... (ii) Again, put x = - 1 in Eq. (i), we get (1−(−1)+(−1)2)n = a0−a1+a2 - ... + a2n ⇒ a0−a1+a2 - ... + a2n = 3n ... (iii) On adding Eq. (ii) and Eq. (iii), we get 2(a0+a2+...+a2n) = 3n + 1 ⇒ a0+a2 + ... + a2n =