<s>[INST] 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?
 [/INST] The answer from lhf has already dealt with the second question posed here.  Here is a Wikipedia article about it.
Here's a probabilistic approach to the first question.  When you toss a coin $n$ times, the probability that a "head" appears exactly $k$ times is $\dbinom n k (1/2)^n$.  The average number of times a "head" appears is therefore
$$
\left( \binom n 0 \cdot 0 + \binom n 1 \cdot 1 + \binom n 2 \cdot 2 + \cdots + \binom n k \cdot k + \cdots + \binom n n \cdot n \right) \left(\frac 1 2 \right)^n.
$$
But the average number of times a "head" appears is obviously $n/2$.  Therefore
$$
\left( \binom n 0 \cdot 0 + \binom n 1 \cdot 1 + \binom n 2 \cdot 2 + \cdots + \binom n k \cdot k + \cdots + \binom n n \cdot n \right) \left(\frac 1 2 \right)^n = \frac n 2.
$$
Consequently
$$
\binom n 0 \cdot 0 + \binom n 1 \cdot 1 + \binom n 2 \cdot 2 + \cdots + \binom n k \cdot k + \cdots + \binom n n \cdot n = n 2^{n-1} .
$$
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