anonymous
  • anonymous
Prove the following proposition using the principle of mathematical induction
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
zepdrix
  • zepdrix
\[\large\rm \sum_{i=1}^n (2i-1)=n^2,\qquad\forall n\in\mathbb Z^+\] Ok so we start with our base case: \(\large\rm n=1\)
zepdrix
  • zepdrix
\[\large\rm (2\cdot 1-1)=2-1=1\]Which is equal to \(\large\rm 1^2\), so our base case is satisfied, ya?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
yes
zepdrix
  • zepdrix
We'll assume it's true for \(\large\rm n=k\). Which means we assume \(\large\rm \color{orangered}{1+3+...+(2k-1)=k^2}\) is true.
zepdrix
  • zepdrix
We call this the Induction Hypothesis ^
anonymous
  • anonymous
yeah
anonymous
  • anonymous
okay
zepdrix
  • zepdrix
Then for our Induction Step: \(\large\rm n=k+1\) Let's see if we can get our formula set up correctly :d
anonymous
  • anonymous
to equal n^2?
zepdrix
  • zepdrix
Well, notice that our last number in the sum will be \(\large\rm k+1\) So we won't end up with \(\large\rm n^2\), we'll end up with \(\large\rm (k+1)^2\) on the right. But setting up the left side is going to be a little tricky. Hopefully you can follow what I'm doing here.
anonymous
  • anonymous
is it not k+2?
zepdrix
  • zepdrix
Your summation is giving you a bunch of terms like this:\[\large\rm (2\cdot\color{royalblue}{1}-1)+(2\cdot\color{royalblue}{2}-1)+...+(2\color{royalblue}{k}-1)+(2\color{royalblue}{(k+1)}-1)=(k+1)^2\]
zepdrix
  • zepdrix
I'll simplify the first few terms like I did the last time.\[\large\rm 1+3+...+(2k-1)+(2(k+1)-1)=(k+1)^2\]
zepdrix
  • zepdrix
k+2? what? :o
anonymous
  • anonymous
i dont get the part (2(k+1)−1)
anonymous
  • anonymous
i dont get where k+1 came from
anonymous
  • anonymous
wait i get it
anonymous
  • anonymous
i get it
zepdrix
  • zepdrix
XD
zepdrix
  • zepdrix
So in the Induction Hypothesis, we let \(\large\rm k\) be the largest number in the sequence. We replaced n by k. In our Induction Step, we're letting \(\large\rm k+1\) be the largest number in the sequence. It's the number that comes after k. So our summation grows a little bit, by one more term. And it should be equivalent to (k+1)^2 now, instead of k^2.
anonymous
  • anonymous
yup
zepdrix
  • zepdrix
If you scroll up, you'll notice I colored our Induction Hypothesis in orange. We need to relate this back to that orange thing.
anonymous
  • anonymous
1+3+...+(2k−1)=k^2
anonymous
  • anonymous
this?
zepdrix
  • zepdrix
Yes, that entire left side is contained within our Induction Step formula.
zepdrix
  • zepdrix
\[\large\rm \color{orangered}{1+3+...+(2k-1)}+(2(k+1)-1)=(k+1)^2\]Do you see it? :)
anonymous
  • anonymous
yeah
zepdrix
  • zepdrix
Since our Induction Hypothesis tells us that the orange stuff is k^2, we can make the switch,\[\large\rm \color{orangered}{k^2}+(2(k+1)-1)=(k+1)^2\]
anonymous
  • anonymous
ohhhhh oh kay
zepdrix
  • zepdrix
From there, you need to turn the left side into (k+1)^2 somehow. Do some Algebra steps expand out the stuff, then try to factor it down.
anonymous
  • anonymous
thank yyou!
zepdrix
  • zepdrix
np \c:/

Looking for something else?

Not the answer you are looking for? Search for more explanations.