If (1+x)15=C0+C1x+C2x2+......+C15x15, then C2+2C3+3C4+......+14C15=
We have (1+x)15=C0+C1x+C2x2+......+C15x15
⇒(1+x)15−1x=C1+C2x+....+C15x14
Differentiating both sides with respect to x, we get
=x.15(1+x)14−(1+x)15+1x2
=C2+2C3x+........+14C15x13
putting x=1, we get
C2+2C3+....+14C15=15.214−215+1=13.214+1.