Prove that $1+4+7+...+(3x-2) = \frac{x(3x-1)}{2}$ for all $x \in N$ by strong induction. Prove that the following holds for $n \in N$ by strong induction 
$$ 1+4+7+...+(3x-2) = \dfrac{x(3x-1)}{2}$$ 
I can see the following for simple (weak) induction. Let $P(x)$ be the proposition for $x$
base case (x=1): 
$$3(1)-2=1$$
and
$$\dfrac{(1)(3(1)-1)}{2}=1$$
$P(1)$ holds true
I.H.: Assume $P(k)$ is true for some $k\geq 1$
I.S.: ($P(k)\implies P(k+1)$) 
$$
1+...+ (3(k+1) -2) = 1 + ... + (3k -2) + (3(k+1)-2)  \\
= \dfrac{k(3k-1)}{2} + (3(k+1)-2) \\
= \dfrac{(3k^2-k)+(6k+2)}{2} \\
= \dfrac{3k^2 + 5k + 2}{2} \\
= \dfrac{(k+1)(3(k+1) -1)}{2} \textit{done!}
$$
But what do I do for strong induction? Would I have different base cases? 
 A: To show something by strong induction, you assume it holds for all cases below $n$ and prove it for $n.$ We might as well assume it for all cases below $k+1$ and prove it for all cases below $k+1$... that's just a relabeling of things. But your proof of $P(k)\to P(k+1)$ is stronger than this since it just assumes $P(k)$ and nothing else below. The only subtlety is for $k=-1$ (i.e. $n=0$) since you can't assume (or even necessarily make sense of) $P(-1),$ so that has to be another case, but you take care of that when you prove $P(0)$ from scratch.
The moral of the story (at least the story I'm telling) is that a proof by weak induction is just a proof by strong induction that doesn't use all the allowed assumptions. So if you've done a proof by weak induction, you've done a proof by strong induction, although sometimes it's superficially easier to think about in weak terms.
All you need to do is take your proof of $P(0)$ and say "I have $P(0),$ so if $P(n)$ is true for all $n<0$ then $P(0)$ holds". And then say "if $n>0$ and $P(k)$ holds for all $k < n,$ then $P(n-1)$ holds, so (insert proof of $P(n-1)\to P(n)$) and thus $P(n)$ holds."
