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

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

2. $\displaystyle\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} +\cdots+\frac{1}{(n-1)\cdot n}$

How do you find the sum of these and prove it by induction? Can someone help me get through this?

For the first one, you're asked to find

$$\sum\limits_{k=0}^n k{n \choose k}$$

Since the binomial coefficients have the $n-k$ symmetry, we can put

$$\sum\limits_{k=0}^n (n-k){n \choose n-k}$$

thus

$$S_n = \sum\limits_{k=0}^n k{n \choose k}=\sum\limits_{k=0}^n (n-k){n \choose n-k}$$

But the RHS is

$$n\sum\limits_{k=0}^n {n \choose n-k}-\sum\limits_{k=0}^n k{n \choose n-k}$$

Now

$$S_n=n\sum\limits_{k=0}^n {n \choose k}-\sum\limits_{k=0}^n k{n \choose k}$$

or

$$S_n=n\sum\limits_{k=0}^n {n \choose k}-S_n$$

$$S_n=n2^n-S_n$$

$$2 S_n=n2^n $$

$$S_n=n2^{n-1} $$

The second one becomes easy once you make use of the telescoping property you've been suggested already.

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 $$ \binom n 0 \cdot 0 + \binom n 1 \cdot 1 + \binom n 2 \cdot 2 + \cdots + \binom n k \cdot k + \cdots + \binom n n \cdot n = n 2^{n-1} . $$