Question

Mathematical induction

Answer 1

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 :)

Answer 2

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

Answer 3

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

Answer 4

wait I'm confused

Answer 5

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

Answer 6

lol, your giving out answers

Answer 7

I'm confused

Answer 8

Me?

Answer 9

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

Answer 10

I thought it was asking to use this to prove this

Answer 11

lol I didn't know I was giving answers

Answer 12

yep it is asking if it is correct

Answer 13

oops

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

on both left and right hand sides of the equation

Answer 19

id say that replacement depends on the structure of the problem

Answer 20

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)

Answer 21

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?

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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)

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

Answer 30

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)

Answer 31

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