`C_(0)+(C_(0)+C_(1))+(C_(0)+C_(1)+C_(2))+...+(C_(0)+C_(1)+...+C_(n))`=

If (1 + x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) + C_(3)x^(3) + …+ C_(n) x^(n) , then C_(0) - (C_(0) - C_(1)) + (C_(0) + C_(1) + C_(2))- (C_(0) + C_(1) + C_(2)+ C_(3)) + ...+ (-1)^(n-1) (C_0) + C_(1) + C_(2) + ...+ C_(n-1)) , when n is even integer is

If C_(0),C_(1),C_(2),........C_(n), are the Binomial coefficients in the expansion of (1+x)^(n),n being even,then C_(0)+(C_(0)+C_(1))+(C_(0)+C_(1)+C_(2))+....+(C_(0)+C_(1)+C_(2)+......+C_(n) is equal to

C_(0)+C_(1)+C_(2)+......+C_(n) =

If C_(0),C_(1),C_(2),…,C_(n) are the binomial coefficients in the expansion of (1+x)^(n)*n being even, then C_(0)+(C_(0)+C_(1))+(C_(0)+C_(1)+C_(2))+….+(C_(0)+C_(1)+C_(2)+…+C_(n-1)) is equal to :

C_(0)C_(1)+C_(1)C_(2)+C_(2)C_(3)+...+C_(n-1)C_(n)

If C_(r) = ""^(n)C_(r) and (C_(0) + C_(1)) (C_(1) + C_(2)) … (C_(n-1) + C_(n)) = k ((n +1)^(n))/(n!) , then the value of k, is

prove that C_(0)C_(n)+C_(1)C_(n-1)+C_(2)C_(n-2)+.........+C_(n)C_(0)=2nC_(n)