Mathematical induction

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Mathematical induction

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

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wait I'm confused
it is also missing a square on the (k+1)
lol, your giving out answers
I'm confused
Me?
what is the question? is it asking if the setup is correct?
I thought it was asking to use this to prove this
lol I didn't know I was giving answers
yep it is asking if it is correct
oops
then we would walk thru the process, to determine if the step they propose is valid.
as freckles pointed out already ... its missing alot of stuff tho
I guess it kind of does make sense as a true false question.
hint: \[\text{ If give } P_n \text{ to find } P_{k+1} \text {just replace the } n \text{ 's with } (k+1)\]
on both left and right hand sides of the equation
id say that replacement depends on the structure of the problem
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)
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?
right but there is also another thing missing on the left hand side
would the left side turn into 2k+8 or something?
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
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
And sorry about taking the fun out of the problem... I didn't realize the question was is this true or not at first.
I thought we were going to prove something. :( And there was just a lot of type-o.
Its okay haha at least i understand it now. Thank you :)
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) \[\color{red}{8+10+12+\cdots+(2k+6)}+(2k+8)=\color{red}{k^2+7k}+(2k+8) \\ \text{ where we want to show the right hand side actually does equal } \\ (k+1)^2+7(k+1)\]
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)
but anyways this is getting into the actual proof of the P_n thing is true for all positive integer n

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