if (1+x)15=C0+C1x+C2x2+…+C15x15 then C2+2C3+3C4+…+14C15 is equal to

we have (1+x)15=C0+C1x+C2x2+…+C15x15⇒ (1+x)15−1x=C1+C2x+…+C15x14On differentiating both sides w.r.t. x, we getx⋅15(1+x)14−(1+x)15+1x2=C2+2C3x+…+14C15x13On putting x = t, we get C2+2C3+…+14C15=15⋅214−215+1=13⋅214+1