If (1+x)n=C0+C1x+C2x2+⋯+Cnxn, then C0C2+C1C3+C2C4+⋯+Cn−2Cn=

(1+x)n=C0+C1x+C2x2+C3x3+⋯+Cn−1xn−1+Cnxn           (1)(x+1)n=C0xn+C1xn−1+C2xn−2+⋯+Cn−1x+Cn                           (2)Multiplying Eqs. (1) and (2) and equating the coefficient of xn−2, we getC0C2+C1C3+C2C4+…+Cn−2Cn= Coefficient of xn−2 in (1+x)2n=2nCn−2=(2n)!(n−2)!(n+2)!