Prove that $1+4+7+...+(3x-2) = \frac{x(3x-1)}{2}$ for all $x \in N$ by strong induction.

Question

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?

@TheDemonix_Hermit Yeah I was pretty sure the weak induction was correct. But the question specifically prompts for strong induction. How would one go about that? — CarterKF, Dec 15 '19 at 05:34
Since you have proved it for base case $1$ , and all the subsequent cases , this answer is okay. — The Demonix _ Hermit, Dec 15 '19 at 05:36
@TheDemonix_Hermit I have googled a bit two and I agree that that are equivalent however my course will not accept a proof by weak induction as strong. So there is some formatting thing I need which I am not really sure how to do. — CarterKF, Dec 15 '19 at 05:37
@spaceisdarkgreen no shame. The course I am taking is riddled with mistakes so this might just one of them. — CarterKF, Dec 15 '19 at 05:42
See these related answer what-exactly-is-the-difference-between-weak-and-strong-induction? , strong-induction-requires-no-base-case? , strong-math-induction-always-requires-more-base-cases-than-common-math-induction? — The Demonix _ Hermit, Dec 15 '19 at 05:43
Replace " Assume P(k) is true for some k≥1 " with " Assume P(n) is true for all $n:1\le n \le k$ for some $k$" and then continue saying exactly what you said. Now you've done strong induction. — fleablood, Dec 15 '19 at 05:50

score 3 · Answer 1 · answered Dec 15 '19 at 05:57

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."

Prove that $1+4+7+...+(3x-2) = \frac{x(3x-1)}{2}$ for all $x \in N$ by strong induction.

1 Answers1