r2Cr=12C1+22C2+32C3+......+n2Cn ∵(1+x)n=C0+C1x+C2x2+C3x3+.....+Cnxn On differentiating both sides w.r.t. x, we get n(1+x)n−1=1.C1+2C2x+3C3x2+.......+nCnxn−1 Now, on multiplying by x both sides, we get nx(1+x)n−1=1.C1x+2C2x2+3C3x3+......+nCnxn Now again on differentiating both sides w.r.t x, we get n[(1+x)n−1+(n−1)x(1+x)n−2] =12C1+22C2x+32C3x2+........+n2Cnxn−1 Put x=1, we get 12⋅C1+22C2+32C3+......+n2Cn =n[2n−1+(n−1)2n−2] =n2n−2[2+n−1]=n(n+1)2n−2