Induction: How to prove propositions with universal quantifiers? In my book, they prove with mathematical induction propositions with successions like this:
$$1 + 3 + 5 + \cdots + (2n-1) = n^2$$
In all exercises. However, recently I took some exercises from a different paper and instead of these it told me to prove this:
$$\forall n \in N (11 / (10^{2n+1} + 1 ))$$
Or perhaps this:
$$ \forall n \in N (n < 2^n) $$
(I can't find how to write the natural numbers set symbol. $N$ is it there.)
And now I'm lost. This is is what I did with the first one:
Prove that the proposition works for $n=1$
$$ 11 / (10^3+1) \implies 11/1001
\implies \exists x \in N(11x=1001)$$
Which is true, if you take $x = 91$.
Assume
$$\forall n \in N (11 / (10^{2n+1} + 1 ))$$
We have to prove:
$$\forall n \in N (11 / (10^{2n+3} + 1 ))$$
We prove it:
Which I don't know how to do. Curiously enough, my book only shows exercises with successions, so I guess that this exercise can be, somehow, written as a succession? I am not sure about that. Any ideas?

You are misunderstanding induction here.
The $\forall$ is always part of it. For example, the real statement for your first result is $$\forall n\in\mathbb N: 1+3+...+(2n-1)=n^2$$
In general, if your equation is: $$\forall n\in\mathbb N: P(n)$$, the principal of mathematical induction says it is enough to show:
$$P(1)$$
and 
$$\forall n\in\mathbb N: P(n)\implies P(n+1)$$
So to prove that $11\mid 10^{2n+1}+1$, you first show $11\mid 10^{3}+1$ and then prove that if $11|10^{2n+1}+1$ then $11|10^{2n+3}+1$.
