Correct option is (b) n.2n–1
\(\therefore\) 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 +...\(\frac n2\)times (Adding the terms equidistant from the begining and the end)
\(= \frac n2 .2^n = n.2^{n-1}\)