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?

If you don't have to use induction, you can do this:
1)
$$
\sum_k k\binom{n}{k} = \sum_k k \frac{n(n-1)\cdots(n-k+1)}{k!} = n\sum_k \binom{n-1}{k-1} = n \sum_k \binom{n-1}{k} = n2^{n-1}
$$
Here $k$ runs over all integers and by convention $\binom{n}{k}$ is defined as zero if $k<0$ or $k > n$. The last equality follows from the fact that $\binom{n-1}{k}$ counts the $k$-element subsets of $\{1,\ldots,n-1\}$, so $\sum_k \binom{n-1}{k}$ counts the number of all subsets which is $2^{n-1}$.
2) Note that $\frac{1}{k(k+1)} = \frac{1}{k}-\frac{1}{k+1}$ so the sum is a telescope sum:
$$\sum_{k=1}^{n-1} \frac{1}{k(k+1)} = \sum_{k=1}^{n-1}\left( \frac{1}{k} - \frac{1}{k+1}\right) = 1 - \frac{1}{n}$$
