MATHEMATICAL INDUCTION

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MATHEMATICAL INDUCTION

Mathematics
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\[1^{2}+4^{2}+7^{2}+....+(3n-2)^{2}=\frac{ n(6n ^{2}-3n-1) }{ 2 }\]
How would I do s(k+1)? @ganeshie8

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Okay thanks that shows me the work but I dont understand it, can you explain to me step by step? @IrishBoy123
sure in inductions, you first prove it is true for some value of n. so establish that bit first.
I already did that part: n=1 and n=k I just dont know how to d the n=k+1
Honestly, I don't get it.
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|
Yeah I have that
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
and the critical step, do you know that too?
that step just confused me im trying to understand it just a min
Question: is it not that for n =k, we have \(....+(3k-2)^2= \dfrac{k(6k^2-3k-1)}{2}\)
|dw:1440080370055:dw|
Woah im so lost now
|dw:1440080425206:dw|
oh okay so 3(k+1-2)^2 takes the place of 6k^2....
yeah
slow down a bit gives me a minute to summarise where we are: will take less than a minute
|dw:1440080585980:dw|
|dw:1440080618737:dw|
so the final step is to show that the RHS is as the original equation would predict ie |dw:1440080671229:dw|
and that's proof done, if we complete that final step
so we now wrestle with the algebra trying to show these 2 are in fact equal.
I did, they are not the same :(
I show you
1 Attachment
Alright I have to go ill be back though sorry You guys can continue helping crazyand... I will catch up with you guys
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
\(\dfrac{k(6k^2-3k-1)}{2}+(3(k+1)-2)^2\\=\dfrac{k(6k^2-3k-1)}{2}+(3k+1)^2\\=\dfrac{k(6k^2-3k-1)}{2}+9k^2+6k+1\\=\dfrac{k(6k^2-3k-1)+18k^2+12k+2)}{2}\\\) \(=\dfrac{6k^3-3k^2-k+18k^2+12k+2}{2}\) \(=\dfrac{6k^3+15k^2+11k+2}{2}\)
right, so far. at this point i divided by k + 1 as we know that is on the RHS if the statement is correct
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. :(
check brackets
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YYYYYYYYYYYES!! I got you. Finally. Thanks for being patient to me.
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.
We learned from each other, right? hehehe..

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