Question

Help with this mathematical induction proof: http://imgur.com/6lALs

@Mathematics

Answer 1

the proof seems to be underway quite nicely, havent checked the math tho

Answer 2

Yes, so now I am not seeing what direction to go next

Answer 3

write out what your "proven" step needs to look like to guide you

Answer 4

an end results

Answer 5

\[\frac{2(k+1)((k+1)+1)(2(k+1)+1)}{3}\]

Answer 6

yes

Answer 7

\[\frac{2(k+1)(k+2)(2k+3)}{3}\]
this is what you need to aim for; you can expand this all out to get to an intermediary level that your proof side will expand to as well

Answer 8

\[\frac{2(k)(k+1)(2k+1)}{3}+(k+1)=expanded.proof.guide\]

Answer 9

\[\frac{2(k)(k+1)(2k+1)}{3}+(k+1)\]
\[\frac{2(k)(k+1)(2k+1)+3(k+1)}{3}\]
we got a 3 on the bottom, thats a good start

\[\frac{(2k^2+2k)(2k+1)+3k+3}{3}\]
\[\frac{4k^3 +2k^2+4k^2+2k+3k+3}{3}\]
\[\frac{4k^3 +6k^2+5k+3}{3}\]
if i did the mathing right

Answer 10

Your beginning is it right?

Answer 11

\[2(k)(k+1)(2k+1)/3+(k+1)\]

Answer 12

Where is this from?

Answer 13

yes, I took the assumed P(k) and added (k+1) to it, but i prolly should have put it into it proper form :) + 2(k+1)^2

Answer 14

I added \[2(k+1)^{2}\]

Answer 15

thats the one, trying to toggle back and forth to see the question makes my life difficult

Answer 16

Alright so what ive gotten with that term added is:

Answer 17

\[(4k^{3}+6k ^{2}+2k)(6k ^{2}+12k+6) / 3\]

Answer 18

Which seems like im getting further and further from what i need

Answer 19

Should I expand it out even more??

Answer 20

hmm

Answer 21

\[\frac{2(k)(k+1)(2k+1)}{3}+2(k+1)^2\]
\[\frac{2(k)(k+1)(2k+1)+6(k+1)^2}{3}\]
\[\frac{2k(2k^2+3+1)+6(k^2+2k+1)}{3}\]
\[\frac{4k^3+6k^2+2k+6k^2+12k+6}{3}\]
\[\frac{4k^3+12k^2+14k+6}{3}\]
expanded proof guide is:

\[\frac{2(k+1)(k+2)(2k+3)}{3}\]
\[\frac{2(k^2+3k+2)(2k+3)}{3}\]
\[\frac{(k^2+3k+2)(4k+6)}{3}\]
\[\frac{4k^3+18k^2+26k+12}{3}\]
the wolf confirms that one:

unless my mathing is off someplace, which could be the case, this is what i get

Answer 22

isnt 2(1)^2 = 2?

Answer 23

would it be?

Answer 24

as is, the case does not hold for P(1)

Answer 25

hmm

Answer 26

is the:
\[\sum2i^2\]or\[\sum(2i)^2\]
???

Answer 27

(2n)^2

Answer 28

then that will make a difference lol

Answer 29

the expansion provided for a proof guide is still good then

Answer 30

So I just gotta add in an extra paren

Answer 31

Didnt the expansions yield different results?

Answer 32

the expansion was the result of 2*i^2, so yes it yeilded different results

Answer 33

\[\frac{2k(k+1)(2k+1)}{3}+[2(k+1)]^2\]
\[\frac{2k(k+1)(2k+1)+3[2(k+1)]^2}{3}\]
\[\frac{(2k^2+2k)(2k+1)+3(4(k+1)^2)}{3}\]
\[\frac{(4k^3+2k^2+4k^2+2k)+12(k^2+2k+1))}{3}\]
\[\frac{4k^3+6^2+2k+12k^2+24k+12}{3}\]
\[\frac{4k^3+18k^2+26k^2+12}{3}\]
whish is the same as the proof guide expansion

Answer 34

so yeah, your err is in adding 2(k+1)^2 instead of (2(k+2))^2

Answer 35

murmur ... typoed the end lol

Answer 36

hmm

Answer 37

Ok, so let me see why did I add the wrong thing...

Answer 38

Ah, ok i see

Answer 39

Sneaky parenthesis for sure!