The coefficient of x5 in the expansion of (1+x)21+(1+x)22+…+(1+x)30 is

(1+x)21+(1+x)22+…+(1+x)30=(1+x)211+(1+x)1+…+(1+x)9=(1+x)21(1+x)10−1(1+x)−1=1x(1+x)31−(1+x)21∴ Coefficient of f in the given expression= Coefficient of x5 in 1x(1+x)31−(1+x)21= Coefficient of x6 in (1+x)31−(1+x)21=31C6−21C6