Correct option (b) n.2n–1
Explanation:
∴ C0 + (C0 + C1 ) +.........+ (C0 + C1 +.......Cn–2) + (C0 + C1 +..........Cn–1)
= (Cn) + (Cn + Cn–1) +.......+ (C0 + C1 +.........+ Cn–2) + (C0 + C1 +.....+ Cn–1)
= 2n + 2n + 2n +... n/2 times (Adding the terms equidistant from the begining and the end)
= n/2 . 2n = n.2n–1