`2C_0+5C_1+8C_2++(3n+2)C_n=(3n+4)2^(n-1)`

If (1+x)^n =C_0+C_1 x+ C_2 x^2 +....... C_nx^n prove the following : 2C_0+5 C_1+8C_2+...+(3n+2)C_n=(3n+4)2^(n-1) .

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

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

3C_0+5C_1+7C_2+ +(2n+3)C_n=2^n(n+3)

Prove the following: 2*C_(0)+3*C_(1)+4*C_(2)+…+(n+2)*C_(n)=(n+4).2^(n-1)

If (1+x)^(n)=C_(0)+C_(1).x+C_(2).x^(2)+….+C_(n).x^(n). then prove that (i) C_(0)+2C_(1)+3C_(2)+…+(n-1)C_(n)=(n+2).2^(n-1) (ii)C_(0)+3C_(1)+5C_(2)+...+(2n+1)C_(n)=(n+1).2^(n)

Show that C_2 + 2C_3 + 3C_4 + ... + (n-1)C_n = 1+ (n-2) 2^(n-1)

C_0 - [C_1 -2.C_2+ 3.C_3-……..+(-1)^(n-1).n.C_n] =

C_0 C_1 + C_1 C_2 + .... + C_(n-1) C_n = (2^n.n.1.3.5... (2n-1))/(n+1)

Show that C_0 + 3C_1 + 5C_2 + .... +(2n+1) C_n = (n+1)(2^n)