We have 2n+1C1+2n+1C2+....+2n+1Cn=255 ....(i) Also, the sum of binomial coefficients 2n+1C0+2n+1C1+2n+1C2+....+2n+1Cn+2n+1Cn+1+2n+1Cn+2+......+2n+1C2n+1 =(1+1)2n+1=22n+1 ⇒ 2n+1C0+2{2n+1C1+2n+1C2+....+2n+1Cn}+2n+1C2n+1=22n+1 ⇒ 1+2(255)+1=22n+1 ⇒ 1+255=22n ⇒ 22n=28 ⇒ n=4