Let (1+x)n=C0+C1x+C2x2+…+Cnxn and C1C0+2C2C1+3C3C2+⋯+nCnCn−1=1kn(n+1), then
value of k is
rCrCr−1=rn!r!(n−r)!(r−1)!(n−r+1)!n!⇒ ∑r=1n rCrCr−1=∑r=1n (n−r+1)=12n(n+1)∴ k=2.