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anonymous
 one year ago
MATHEMATICAL INDUCTION
anonymous
 one year ago
MATHEMATICAL INDUCTION

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[1^{2}+4^{2}+7^{2}+....+(3n2)^{2}=\frac{ n(6n ^{2}3n1) }{ 2 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How would I do s(k+1)? @ganeshie8

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1see here http://openstudy.com/users/irishboy123#/updates/55d3474fe4b0554d62727bc4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay thanks that shows me the work but I dont understand it, can you explain to me step by step? @IrishBoy123

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1sure in inductions, you first prove it is true for some value of n. so establish that bit first.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I already did that part: n=1 and n=k I just dont know how to d the n=k+1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Honestly, I don't get it.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1OK, 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

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1i 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
 one year ago
Best ResponseYou've already chosen the best response.1and the critical step, do you know that too?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that step just confused me im trying to understand it just a min

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Question: is it not that for n =k, we have \(....+(3k2)^2= \dfrac{k(6k^23k1)}{2}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440080370055:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1440080425206:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh okay so 3(k+12)^2 takes the place of 6k^2....

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1slow down a bit gives me a minute to summarise where we are: will take less than a minute

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1440080585980:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1440080618737:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1so the final step is to show that the RHS is as the original equation would predict ie dw:1440080671229:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1and that's proof done, if we complete that final step

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1so we now wrestle with the algebra trying to show these 2 are in fact equal.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I did, they are not the same :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright I have to go ill be back though sorry You guys can continue helping crazyand... I will catch up with you guys

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1that'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
 one year ago
Best ResponseYou've already chosen the best response.0\(\dfrac{k(6k^23k1)}{2}+(3(k+1)2)^2\\=\dfrac{k(6k^23k1)}{2}+(3k+1)^2\\=\dfrac{k(6k^23k1)}{2}+9k^2+6k+1\\=\dfrac{k(6k^23k1)+18k^2+12k+2)}{2}\\\) \(=\dfrac{6k^33k^2k+18k^2+12k+2}{2}\) \(=\dfrac{6k^3+15k^2+11k+2}{2}\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1right, so far. at this point i divided by k + 1 as we know that is on the RHS if the statement is correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0while other is \(\dfrac{(k+1)(6(k+1)^23(k+1)1}{2}\) \(=\dfrac{(k+1)(6(k^2+2k+1))3k31}{2}\) \(\dfrac{(k+1)(6k^2+12k+6)3k4}{2}\) \(=\dfrac{6k^3+12k^2+6k+6k^2+12k+63k4}{2}\) \(=\dfrac{6k^3+18k^2+15k+2}{2}\) Which is different from the previous one. :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0YYYYYYYYYYYES!! I got you. Finally. Thanks for being patient to me.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1thank 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
 one year ago
Best ResponseYou've already chosen the best response.0We learned from each other, right? hehehe..
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