## anonymous one year ago Mathematical induction

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1. 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 :)

2. 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

3. freckles

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

4. freckles

wait I'm confused

5. freckles

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

6. amistre64

7. anonymous

I'm confused

8. freckles

Me?

9. amistre64

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

10. freckles

I thought it was asking to use this to prove this

11. freckles

lol I didn't know I was giving answers

12. anonymous

yep it is asking if it is correct

13. freckles

oops

14. amistre64

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

15. amistre64

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

16. freckles

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

17. freckles

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

18. freckles

on both left and right hand sides of the equation

19. amistre64

id say that replacement depends on the structure of the problem

20. 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)

21. 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?

22. freckles

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

23. anonymous

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

24. 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

25. 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

26. 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.

27. freckles

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

28. anonymous

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

29. 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) $\color{red}{8+10+12+\cdots+(2k+6)}+(2k+8)=\color{red}{k^2+7k}+(2k+8) \\ \text{ where we want to show the right hand side actually does equal } \\ (k+1)^2+7(k+1)$

30. 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)

31. freckles

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