Prove that $4 + 5 + 6 + \cdots+ n = \frac{n(n+1)}{3}$ for $n \ge 4$ 
Prove using mathematical induction that $$4 + 5 + 6 + \cdots + n = \frac{n(n+1)}{3}$$ where $n \geq 4$ is an integer.

I just wanted to confirm because my Base case P(4) is false. So this statement can't be proven?
 A: Let´s think a little bit about this:
We know $1+...+n= \frac{n(n+1)}{2}$ for evey $n \in \mathbb{N}$. So if $4+5+...+n=\frac{n(n+1)}{3}$ for some $n$, we would have 
$(1+...+n)-(4+....+n)= \frac{n(n+1)}{2}-\frac{n(n+1)}{3} = 1+2+3=6$
And therefore $n(n+1)=36$. 
So we have proven that this is false for every $n \in \mathbb{N}$
A: For $n=4$,
$$4=\frac{4\cdot5}3$$
is blatantly false, hence the claim is wrong. Full stop.
A: A good principle to go by is "Everything is unknown, until proven or taught otherwise." Another principle is "if something is unknown, then you cannot use it".  And a third principle is "if you cannot use something, try something else".
Induction says that P(n) will always be true if the base case is true and the induction step is true.  What about if the base case is false, well it is unknown whether induction would work, hence we may not use it, hence we should try something else.
Now the eventually goal of such a question is to show that $P$ is always true.  So you wanted to show that $P(4)$ is true, and $P(5)$ is true ect.  You have shown that $P(4)$ is false, hence you have proven that $P$ cannot always be true.
