(i)$\binom{n}{1}+2\cdot\binom{n}{2}+3\cdot\binom{n}{3}+...+n\cdot\binom{n}{n}=n\cdot2^{n-1}$
(ii)$\binom{n}{1}+2^2\cdot\binom{n}{2}+3^2\cdot\binom{n}{3}+...+n^2\cdot\binom{n}{n}=n(n+1)\cdot2^{n-2}$
I assume that the binomial series can maybe prove the formulas above $(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+...+\binom{n}{n}x^n$
$$f_n(x)=(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+...+\binom{n}{n}x^n$$ $$f_n'(x)=n(1+x)^{n-1}=\binom{n}{1}+2x\cdot\binom{n}{2}+3x^2\cdot\binom{n}{3}+...+nx^{n-1}\cdot\binom{n}{n}$$
Evaluate everything for $x=1$ and you have the answer to (i).