anonymous
  • anonymous
MATHEMATICAL INDUCTION
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[1^{2}+4^{2}+7^{2}+....+(3n-2)^{2}=\frac{ n(6n ^{2}-3n-1) }{ 2 }\]
anonymous
  • anonymous
How would I do s(k+1)? @ganeshie8

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

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