The AM of  2n+1C0,2n+1C1,2n+1C2,…,2n+1Cn is

2n+1C0+2n+1C1+2n+1C2+…+2n+1C2n+2n+1C2n+1=22n+1Now  2n+1C0=2n+1C2n+1, 2n+1C1=2n+1C2n…2n+1Cr=2n+1C2n−r+1So, sum of first (n + 1 ) terms= sum of last (n + 1 ) terms⇒ 2n+1C0+2n+1C1+2n+1C2+…+2n+1Cn=22n⇒  2n+1C0+2n+1C1+2n+1C2+…+2n+1Cnn+1=22n(n+1)