If (1+x)n=C0+C1x+C2x2+…+Cnxn  then C02+C12+C22+C32+…+Cn2 is equal to

we have ,(1+x)n=C0+C1x+C2x2+…+Cnxn-----(i) and 1+1xn=C0+C11x+C21x2+…+Cn1xn…(ii)on multiplying Eqs. (i) and (ii) and taking thecoefficient of constant terms in right hand side=C02+C12+C22+…+Cn2In right hand side (1+x)n1+1xn or in 1xn(1+x)2n or term containing xn in (1 + x)2n.clearly, the coefficient of xn in (1 + x)2n is equal to 2nCn=(2n)!n!n!