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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

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.

Yeah I have that

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|

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

check brackets

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

We learned from each other, right? hehehe..