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) Take $\displaystyle f(x)= \sum_{i=0}^n\binom{n}{i} x^i=(x+1)^n$. Now consider $f'(1)$.
2) Use that $\frac{1}{(k-1)\cdot k} = \frac{1}{k-1} -\frac{1}{k}$.
