anonymous
  • anonymous
Mathematical induction
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • 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 :)
amistre64
  • 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
  • freckles
that one line seems to be missing +2(k+1)+6 on the left hand side

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freckles
  • freckles
wait I'm confused
freckles
  • freckles
it is also missing a square on the (k+1)
amistre64
  • amistre64
lol, your giving out answers
anonymous
  • anonymous
I'm confused
freckles
  • freckles
Me?
amistre64
  • amistre64
what is the question? is it asking if the setup is correct?
freckles
  • freckles
I thought it was asking to use this to prove this
freckles
  • freckles
lol I didn't know I was giving answers
anonymous
  • anonymous
yep it is asking if it is correct
freckles
  • freckles
oops
amistre64
  • amistre64
then we would walk thru the process, to determine if the step they propose is valid.
amistre64
  • amistre64
as freckles pointed out already ... its missing alot of stuff tho
freckles
  • freckles
I guess it kind of does make sense as a true false question.
freckles
  • freckles
hint: \[\text{ If give } P_n \text{ to find } P_{k+1} \text {just replace the } n \text{ 's with } (k+1)\]
freckles
  • freckles
on both left and right hand sides of the equation
amistre64
  • amistre64
id say that replacement depends on the structure of the problem
amistre64
  • 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
  • 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
  • freckles
right but there is also another thing missing on the left hand side
anonymous
  • anonymous
would the left side turn into 2k+8 or something?
freckles
  • 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
  • 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
  • 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
  • freckles
I thought we were going to prove something. :( And there was just a lot of type-o.
anonymous
  • anonymous
Its okay haha at least i understand it now. Thank you :)
freckles
  • 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
  • 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
  • freckles
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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