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

Let E=C0+2C1+3C2+…+nCn−1+(n+1)Cn            (1)Using Cr=Cn-r, we can rewrite (1) asE=(n+1)C0+nC1+(n−1)C2+…+2Cn-1+Cn              (2)Adding (1) and (2), we get                         2E=(n+2)C0+(n+2)C1+(n+2)C2+…                    +(n+2)Cn                =(n+2)C0+C1+…+Cn=(n+2)2n⇒          E=(n+2)2n−1Alternative SolutionWe have C0x+C1x2+C2x3+…+Cnxn+1=x(1+x)nDifferentiating both the sides, we get             C0+2C1x+3C2x2+…+(n+1)Cnxn=(1+x)n+nx(1+x)n−1      (1)Putting x=1, we get          C0+2C1+3C2+…+(n+1)Cn=2n+n(1)2n−1=(n+2)2n−1