If ${\mathrm{C}}_0,{\mathrm{C}}_1,{\mathrm{C}}_2,\dots ,{\mathrm{C}}_n$ denote the binomial coefficients in the expansion of $(1+x{)}^n$ then $1^3\cdot C_1+2^3\cdot C_2+3^3\cdot C_3+\dots +n^3\cdot C_n=$

detailed solution

Correct option is B

We have,13⋅C1+23⋅C2+33⋅C3+…+n3⋅Cn=∑r=1n r3⋅Cr=∑r=1n r3⋅nCr=∑r=1n [r(r−1)(r−2)+3r(r−1)+r]nCr=∑r=1n r(r−1)(r−2)nCr+∑r=1n 3r(r−1)nCr+∑r=1n r⋅nCr=∑r=3n r(r−1)(r−2)⋅nr⋅n−1r−1⋅n−2r−2⋅n−3Cr−3+∑r=2n 3r(r−1)nr⋅n−1r−1n−2Cr−2+∑r=1n r⋅nrn−11Cr−1=n(n−1)(n−2)∑r=3n n−3Cr−3+3n(n−1)∑r=2n n−2nCr−2+n∑r=1n n−1Cr−1=n(n−1)(n−2) n−3C0+n−3C1+…+n−3Cn−3+3n(n−1) n−2C0+n−2C1+…+n−2Cn−2+n n−1C0+n−1C1+…+n−1Cn−1=n(n−1)(n−2)⋅2n−3+3n(n−1)⋅2n−2+n⋅2n−1={(n−1)(n−2)+6(n−1)+4}n2n−3=nn2+3n2n−3=n2(n+3)2n−3