Prove that `C_(0)+3.C_(1)+5.C_(2)+….+(2n+1).C_(n)=(2n+2).2^(n-1).`

Prove that : Prove that 2.C_(0)+7.C_(1)+12.C_(2)+……..+(5n+2)C_(n)=(5n+4)2^(n-1).

Prove the C_(0)+2.C_(1)+4.C_(2)+8.C_(3)+…+2^(n).C_(n)=3^(n)

Prove the following: C_(0)+2*C_(1)_+3*C_(2)+…(n+1)*C_(n)=(n+2).2^(n-1)

Prove that : Prove that C_(0)+3.C_(1)+3_(2).C_(2)+……+3^(n).C_(n)=4^(n)

Prove that following i) 2.C_(0)+5.C_(1)+8.C_(2)+…..+(3n+2)C_(n)=(3n+4).2^(n-1)

Prove that (C_(0) +C_(1)+C_(2)+….+C_(n))^(2)=1 +""^(2n)C_(1) +""^(2n)C_(2) +…..+""^(2n)C_(2n)

With usual notations prove that 2.C_0 + 7.C_1 + 12.C_2 + ……..+(5n + 2).C_n = (5n + 4).2^(n-1)

Prove that : If n is a positive integer, then prove that C_(0)+(C_(1))/(2)+(C_(2))/(3)+….+(C_(n))/(n+1)=(2^(n+1)-1)/(n+1).

Prove that : For n = 0, 1, 2, 3, ………., n, prove that C_(0).C_(r)+C_(1).C_(r+1)+C_(2).C_(r+2)+….+C_(n-r).C_(n) =""^(2n)C_((n+r)) and hence deduce that Prove that : C_(0)^(2)+C_(1)^(2)+C_(2)^(2)+….+C_(n)^(2)=""^(2n)C_(n)

Prove that C _(0) +2 . C _(1) + 2 ^(2) . C _(2) + .......+ 2 ^(n) . C _(n) = 3 ^(n).