Evaluate and prove by induction: $\sum k{n\choose k},\sum \frac{1}{k(k+1)}$ 
*

*$\displaystyle
  0\cdot \binom{n}{0} + 1\cdot \binom{n}{1} + 2\binom{n}{2}+\cdots+(n-1)\cdot \binom{n}{n-1}+n\cdot \binom{n}{n}$

*$\displaystyle\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} +\cdots+\frac{1}{(n-1)\cdot n}$
How do you find the sum of these and prove it by induction? Can someone help me get through this?

1)
$$S_n=\sum_{k=0}^nk\frac{n!}{k!(n-k)!}=n\sum_{k=1}^n\frac{(n-1)!}{(k-1)!(n-k)!}=n\sum_{k=0}^{n-1}\binom{n-1}k=n2^{n-1}.$$
Then the inductive proof by $S_0=0,S_n-S_{n-1}=T_n$, is a little tedious:
$$S_n-S_{n-1}
=\sum_{k=0}^nk\binom nk-\sum_{k=0}^{n-1}k\binom {n-1}k=n\binom nn+\sum_{k=0}^{n-1}k\left(\binom nk-\binom{n-1}k\right)\\
=n+\sum_{k=1}^{n-1}k\binom{n-1}{k-1}
=n+\sum_{k=1}^{n-1}(k-1)\binom{n-1}{k-1}+\sum_{k=1}^{n-1}\binom{n-1}{k-1}\\
=n+\sum_{k=1}^{n-1}(k-1)\binom{n-1}{k-1}+\sum_{k=1}^{n-1}\binom{n-1}{k-1}\\
=n+\sum_{k=0}^{n-2}k\binom{n-1}k+\sum_{k=0}^{n-2}\binom{n-1}{k}\\
=n+S_{n-1}-(n-1)+2^{n-1}-1=S_{n-1}+2^{n-1}.$$
And on the other hand
$$T_n=S_n-S_{n-1}=n2^{n-1}-(n-1)2^{n-2}=(n-1)2^{n-2}+2^{n-1}$$
2)
As $\dfrac1{k(k+1)}=\dfrac1k-\dfrac1{k+1}$, the sum telescopes as
$$S_n=\frac1{n+1}-1.$$
This time, the proof by induction is trivial:
$$S_0=0,$$
$$T_n=S_n-S_{n-1}=\frac1{n+1}-1-\frac1n+1=\frac1{n(n+1)}.$$
