If(1+x)n=C0+C1x+C2x2+…+Cnxn them the value of C0+2C1+3C2+…+(n+1)Cn will be

Since, x(1+x)n=xC0+C1x2+C2x3+…+Cnxn+1 On differentiating w.r.t. x, we get (1+x)n+nx(1+x)n−1=C0+2C1x+3C2x2+…+(n+1)CnxnPut x = L, we get C0+2C1+3C2+…+(n+1)Cn=2n+n2n−1=2n−1(n+2)