Can mathematical induction be used to disprove something? I saw this to be the rule of inference for mathematical induction :

Now consider :
 as L.H.S.
and 
  as R.H.S.. 
Now if suppose, while trying to prove P(k) -> P(k+1), in the left hand side of the expression, comes out to be false for some case, then the whole left hand side becomes false.
But A -> B just imposes a condition that whenever A is true, B must also be true. There is no boundation on B when A is false.
That means even if my LHS is False, RHS could be True!.
And hence, Mathematical induction can only be used to prove something, and not disprove ?
Is my argument correct ? If someone could throw more light please ?

Edit:
What I mean to say is this :
Suppose I come up with a formula for something and want to prove that this formula is correct. I apply mathematical induction and LHS comes out to be False. My formula can still be correct, because it doesnt depend on the Truth false of LHS when LHS itself is false, just that Mathematical induction is not able to prove it ?
 A: (It's not clear to me what you are asking, and your question could be clarified slightly.)
Here's an example for you to consider, which might be relevant to what you are thinking.
Suppose we want to find when $n! \geq 3^n$.
Now, assume it is true for some $k$. Then, if $k + 1 \geq 3$, we can apply the induction hypothesis to see that
$$(k+1)! = (k+1) \times k! \geq (k+1) \times 3^k \geq 3^{k+1}$$
However, this is not true for $n=2, 3, 4, 5, 6$. But it is true for $n=7$ (and thereafter).
Hence, we have a case where
1. P(6) is not true,
2. if P(6) is true then P(7) is true,
3. P(7) is true.
A: I'm going to answer the question in the body, which seems to me to be different than the answer in the title.
If the LHS is false, then either $P(0)$ is false (the base case fails) or 
there is a natural number $k$ such that the implication $P(k) \implies P(k+1)$ is false (the induction step fails.)
If $P(0)$ is false, then $0$ is a counterexample to the universal statement $\forall n \, P(n)$.
If the implication $P(k) \implies P(k+1)$ is false, then this means that $P(k)$ holds but $P(k+1)$ fails. Therefore the natural number $n = k+1$ is a counterexample to the universal statement $\forall n \, P(n)$.
Therefore in either case the RHS, "$\forall n \, P(n)$," fails.
What this shows is that any universal statement about natural numbers can be proved by induction.  However, this is only true in a trivial sense: it could be that proving the implication $P(k) \implies P(k+1)$ for general $k$ is no easier than proving the conclusion $P(k+1)$ for general $k$.
