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)`

Explanation

LHS. .`=underset(r=1)overset(n)Sigma r.""^(n)C_(r)==underset(r=1)overset(n)Sigma r.""^(n-1)C_(r-1)`
`underset(r=1)overset(n)Sigmar.n/r.""^(n-1)C_(r-1)`
`n=underset(r=1)overset(n)SigmaC_(r-1)=n[""^(n-1)C_(0)+""^(n=1)C_(1)+.....+""^(n+1)C_(n+1)]`
`=n.2^(n-1)`
Aliter :(using method of differentiation )
Intergrating (A) we get
`=((1+x)^(n) = ""^(n)C_(0)+""^(C)_(2)x^(2)+.....+ ""^(n)C_(n)x^(n)`
Differentitating (A) we get
` n(1+x)^(n+1)=C_(1)+2C_(2)x+3C_(3)X^2+...+n.C_(n)x^(n+1)`
Put X=1
`C_(1)+2C_(2)+3C_(3)+...+n.C_(n)=n.2^(n-1)`
(ii) ` L.H.S = underset(r=0)overset(n)Sigma(C_(r))/(r+1)=(1)/(n+1)underset(r=0)overset(n)Sigma (n+1)/(r+1)""^(n)C_(r)`
`=(1)/(n+1)underset(r=0)overset(n)Sigma^(n+1)C_(r+1)=(1)/(n+1)[""^(n+1)C_(1)+""^(n+1)C_(2)+......+""^(n+1)C_(n+1)]=(1)/(n+1)]=(1)/(n+1)[2^(n+1)-1]`
`((1+x)^(n+1))/(n+1)+C=C_(0)x+(C_(1)x^2)/(2)+(C_(2)x^3)/(3)+.....+(C_(n)x^(n+1))/(n+1)` (where C is a constant )
put =0 ,we get `C=-(1)/(n+1)`
`therefore ((1+x)^(n+1)-1)/(n+1)=C_(0)x+(C_(1)x^2)/(2)+(C_(2)x^3)/(3)+....+(C_(n)x^(n+1))/(n+1)`
Put x=1,we get `C_(0)(C_(1))/(2)+C_(2)/3+....+C_(n)/(n+1)=(2^(n+1)-1)/(n+1)`
Put x=1,we get `C_(0)(C_(1))/(2)+C_(2)/3+....=(1)/(n+1)`