Mathematical induction

- anonymous

Mathematical induction

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

Can someone help me with this? Don't just give me the answer but i need the steps too cuz i don't understand it. Thanks :)

##### 1 Attachment

- amistre64

well we should start with the first step ... is it true for some value of n?
are we restricted by our n values? these usually define a domain

- freckles

that one line seems to be missing +2(k+1)+6 on the left hand side

Looking for something else?

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

## More answers

- freckles

wait I'm confused

- freckles

it is also missing a square on the (k+1)

- amistre64

lol, your giving out answers

- anonymous

I'm confused

- freckles

Me?

- amistre64

what is the question? is it asking if the setup is correct?

- freckles

I thought it was asking to use this to prove this

- freckles

lol I didn't know I was giving answers

- anonymous

yep it is asking if it is correct

- freckles

oops

- amistre64

then we would walk thru the process, to determine if the step they propose is valid.

- amistre64

as freckles pointed out already ... its missing alot of stuff tho

- freckles

I guess it kind of does make sense as a true false question.

- freckles

hint:
\[\text{ If give } P_n \text{ to find } P_{k+1} \text {just replace the } n \text{ 's with } (k+1)\]

- freckles

on both left and right hand sides of the equation

- amistre64

id say that replacement depends on the structure of the problem

- amistre64

the proofing in this case requires us to add the (k+1)th term to each side and see if it simplifies to a format that looks as if all we did was replace n by (k+1)

- anonymous

Oh wait so since when you substitute the n's for (k+1) the top equation has a k^2 and the bottom doesn't so it already wouldn't be equal?

- freckles

right but there is also another thing missing on the left hand side

- anonymous

would the left side turn into 2k+8 or something?

- freckles

2(k+1)+6
2k+2+6
2k+8
yep that should be the last term on the left hand side
where the one before it was 2k+6

- freckles

you know because if it was 8+10+12+...+(2k+6)
which is the exact thing as 8+10+12+...+(2n+6)
except we have k instead of n
then we should have this is k^2+7k since that other one is n^2+7n

- freckles

And sorry about taking the fun out of the problem...
I didn't realize the question was is this true or not at first.

- freckles

I thought we were going to prove something. :(
And there was just a lot of type-o.

- anonymous

Its okay haha at least i understand it now. Thank you :)

- freckles

and yeah the way we prove it is true for k+1
is take that previous thing replace n's with k's
then as @amistre64 said had the (k+1)th term on both sides
which you already know is +(2k+8)
\[\color{red}{8+10+12+\cdots+(2k+6)}+(2k+8)=\color{red}{k^2+7k}+(2k+8) \\ \text{ where we want to show the right hand side actually does equal } \\ (k+1)^2+7(k+1)\]

- freckles

and if you are doing that you might have an easier time expanding (k+1)^2+7(k+1) to show it is the same as k^2+7k+(2k+8)

- freckles

but anyways this is getting into the actual proof of the P_n thing is true for all positive integer n

Looking for something else?

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