C0+2C1+3C2+4C3+...........+(n+1)Cn
So, we know that
(1+x)2=C0+C1x+C2x2+C3x3+C4x4+.......+Cnxn−−−−(1)
n.(1+n)n−1=0+C1+2C2x+3C3x2+.........+nCnx(n−1)
diff. with respect to x
n.(1+n)n−1=0+C1+2C2x+3C3x2+.........+nCnx(n−1)
put x=1 in eq(1) and eq(2)
2n+2(2n−1)=C0+2C1+3C2+4C3+.......+(n+1)Cn
2n+n.2n−1 Required value