Use induction and Newton's binomial formula to show that $\binom{n}{0}+\binom{n}{1}+\cdot+\binom{n}{n}=2^n, \forall n\in \mathbb N$ Use induction and Newton's binomial formula to show that:
$ i)$ $ \binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}=2^n, \forall n\in \mathbb N$
$ ii)$ $\binom{n}{0}-\binom{n}{1}+\binom{n}{3}-\cdots+(-1)^n\binom{n}{n}=0, \forall n\in \mathbb N$
$ iii)$ $\binom{n}{1}+2\binom{n}{2}+3\binom{n}{3}+\cdots+n\binom{n}{n}=n(2^{n-1}), \forall n\in \mathbb N$
$ iv)$ $1\cdot 2\binom{n}{2}+2\cdot 3\binom{n}{3}+\cdots+(n-1)\cdot n\binom{n}{n}=n(n-1)2^{n-2}, \forall n\in \mathbb N$
I really don't know how I can start with this.
 A: This feels like your homework and I'm not comfortable simply solving you the problems. I'm not sure if you already know that, but I will simply explain and provide examples of the needed subjects. You can report any specific problems you are having:
Proof by induction
You can use that to proof statements about natural numbers. Identify the statement $P(n)$ you wish to prove and do the following:
1) Prove the base case, ie, the smallest natural number for which the statement is true.
2) Suppose P(n) is true and prove that it implies in P(n+1) being true.
Example: $1 + 2 + 3 + ... + n = \frac{(n)(n+1)}{2}$
It is clearly true for n = 0. Supposing it is true for n, we have:
$1 + 2 + 3 + ... + n + (n+1) = \frac{(n)(n+1)}{2} + (n+1) = \frac{(n)(n+1) + 2(n+1)}{2} = \frac{(n+1)(n+2)}{2}$
Which is the statemente for $n+1$. This ends the proof.
Newton's binomial formula
I won't prove it. I will simply show and use one or two times.
$(a + b)^n = \sum_{i=0}^{n}  a^ib^{n-i}\binom{n}{i}$
For example:
$1.04^2 = (1 + 0.04)^2 = 1^0\times0.04^2 \binom{2}{0} + 1^1\times0.04^1 \binom{2}{1} + 1^2\times0.04^0 \binom{2}{2} = 1.0816$
Of course this is a very complicated way of doing something simple. But the formula should help in several of your exercices. Try to apply for the correct values of $a$ and $b$.
A: Here is an alternative approach (without Newton's Binomial):
Let's take an $n$-bit zero binary string. There are $\binom{n}{k}$ ways to choose $k$ bits and turning them on. Since the total number of ways of choosing how to turn them on is $2^n$, but at the same time is the number of ways of turning $0,1,2,...n$ bits on, we are done.
