In the expansion of (1 + x)n ,C1C0+2C2C1+3C3C2+…+nCnCn−1 is equal to
(n+1)2
n2
n(n+1)2
None of these
Since, we know that nCr nCr−1=n−r+1r
∴C1C0=n,C2C1=n−12,C3C2=n−23,…
Thus, C1C0+2C2C1+3C3C2+…=n+2⋅n−12+3⋅n−23+…+n⋅1n
=[n+(n−1)+(n−2)+…+1]=n(n+1)2