MATHEMATICAL INDUCTION

- anonymous

MATHEMATICAL INDUCTION

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- anonymous

\[1^{2}+4^{2}+7^{2}+....+(3n-2)^{2}=\frac{ n(6n ^{2}-3n-1) }{ 2 }\]

- anonymous

How would I do s(k+1)? @ganeshie8

- IrishBoy123

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## More answers

- anonymous

Okay thanks that shows me the work but I dont understand it, can you explain to me step by step? @IrishBoy123

- IrishBoy123

sure
in inductions, you first prove it is true for some value of n. so establish that bit first.

- anonymous

I already did that part: n=1 and n=k I just dont know how to d the n=k+1

- anonymous

Honestly, I don't get it.

- IrishBoy123

OK, one important thing. last time i helped on this the OP specified "for all positive integers n" so yes you had to start with n = 1. we will use induction to prove that it must therefore be true for n = 2, and thuse n = 3 and thus ....
so the next step is to assume it is true for n = k. that means the following is assumed true
|dw:1440079974650:dw|

- anonymous

Yeah I have that

- IrishBoy123

i follows therefore that :
|dw:1440080053278:dw|
that has to be true as we have added the same amount to each side
as it happens what we added to LHS was next term in the series

- IrishBoy123

and the critical step, do you know that too?

- anonymous

that step just confused me im trying to understand it just a min

- anonymous

Question: is it not that for n =k, we have \(....+(3k-2)^2= \dfrac{k(6k^2-3k-1)}{2}\)

- anonymous

|dw:1440080370055:dw|

- anonymous

Woah im so lost now

- IrishBoy123

|dw:1440080425206:dw|

- anonymous

oh okay so 3(k+1-2)^2 takes the place of 6k^2....

- anonymous

yeah

- IrishBoy123

slow down a bit
gives me a minute to summarise where we are:
will take less than a minute

- IrishBoy123

|dw:1440080585980:dw|

- IrishBoy123

|dw:1440080618737:dw|

- IrishBoy123

so the final step is to show that the RHS is as the original equation would predict
ie
|dw:1440080671229:dw|

- IrishBoy123

and that's proof done, if we complete that final step

- IrishBoy123

so we now wrestle with the algebra trying to show these 2 are in fact equal.

- anonymous

I did, they are not the same :(

- anonymous

I show you

- IrishBoy123

##### 1 Attachment

- anonymous

Alright I have to go ill be back though sorry You guys can continue helping crazyand... I will catch up with you guys

- IrishBoy123

that's all you need to know, the rest is the algebra. it's turgid, that's why i shared my working -- which shows one approach

- anonymous

\(\dfrac{k(6k^2-3k-1)}{2}+(3(k+1)-2)^2\\=\dfrac{k(6k^2-3k-1)}{2}+(3k+1)^2\\=\dfrac{k(6k^2-3k-1)}{2}+9k^2+6k+1\\=\dfrac{k(6k^2-3k-1)+18k^2+12k+2)}{2}\\\)
\(=\dfrac{6k^3-3k^2-k+18k^2+12k+2}{2}\)
\(=\dfrac{6k^3+15k^2+11k+2}{2}\)

- IrishBoy123

right, so far. at this point i divided by k + 1 as we know that is on the RHS if the statement is correct

- anonymous

while other is
\(\dfrac{(k+1)(6(k+1)^2-3(k+1)-1}{2}\)
\(=\dfrac{(k+1)(6(k^2+2k+1))-3k-3-1}{2}\)
\(\dfrac{(k+1)(6k^2+12k+6)-3k-4}{2}\)
\(=\dfrac{6k^3+12k^2+6k+6k^2+12k+6-3k-4}{2}\)
\(=\dfrac{6k^3+18k^2+15k+2}{2}\)
Which is different from the previous one. :(

- IrishBoy123

check brackets

##### 1 Attachment

- anonymous

YYYYYYYYYYYES!! I got you. Finally. Thanks for being patient to me.

- IrishBoy123

thank you for showing me a shorter proof. i don't know what the hell i was doing using long division last time i did this with someone .... just plain nuts.

- anonymous

We learned from each other, right? hehehe..

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