`C_(1)+C_(2)+C_(3)+....C_(n)=2^(n)-1`

Show that C_(1)+C_(2)+C_(3)+...+C_(n)=1+2+2^(2)+...+2^(n-1)

Prove that (i) C_(1)+2C_(2)+3C_(3)+……+nC_(n)=n.2^(n+1) (ii) C_(0)+(C_(1)/(2)+(C_(2))/(3)+….+(C_(n))/(n+1)=(2^(n+1)-1)/(n+1)

If (1+x)^(n)=C_(0)+C_(1).x+C_(2).x^(2)+C_(3).x^(3)+......+C_(n).x^(n), then prove that C_(0)+2C_(1)+4C_(2)+6C_(3)+...+2n.C_(n)=1+n*2^(n)

(C_(0)+C_(1))(C_(1)+C_(2))(C_(2)+C_(3))...(C_(n-1)+C_(n))=(C_(0).C_(1).C_(2)......C_(n-1)(n+1)^(n))/(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 (1+a)^(n)=.^(n)C_(0)+.^(n)C_(1)a+.^(n)C_(2)a^(2)+ . . +.^(n)C_(n)a^(n) , then prove that .^(n)C_(1)+2.^(n)C_(2)+3.^(n)3C_(3)+ . . .+n.^(n)C_(n)=n.2^(n-1) .

If (1+x)^(n)=C_(0)+C_(1)x+C_(2)x^(2)+...+C_(n)x^(n) then 2C_(0)+(C_(1))/(2)+2^(3)(C_(2))/(3)+...+2^(n+1)(C_(n))/(n+1)=

If quad (1+x)^(n)=C_(0)+C_(1)x+C_(2)x^(2)+...+C_(n)x^(n), then C_(0)C_(2)+C_(1)C_(3)+C_(2)C_(4)+...+C_(n-2)C_(n)=((2n)!)/((n!)^(2)) b.((2n)!)/((n-1)!(n+1)!) c.((2n)!)/((n-2)!(n+2)!) d.none of these