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I suggest writing out that second half of the equation in terms like the first. ...=n(6n^2-3n-1)/2

I know the statement for n=1 is true but I don't know how to find if k+1 is true

Just check the equation again using the next integer, which would be 2

At that point what would I do next though?

Sorry that it got cut off, its just a -1 in the numerator that got cut off.

Thus we can do this substitution:
\[\frac{ k(6k^{2} -3k-1) }{ 2 } + [3(k+1)-2]^{2} \]

Now just to do the algebra and try to get it to look like k+1

\[\frac{ k(6k^{2} - 3k-1) }{ 2 }+ [3(k+1)-2]^{2} = \frac{ k(6k^{2} - 3k - 1)+2(3k+1)^{2} }{ 2 }\]

why do you add on the + [3(k+1)-2]^2

Oh alright that makes sense

Ill let you finish first

Bleh, lost what I was typing up, so gotta retype it, hang on

no worries thank you so much!

Ive been trying to figure this out for a while now thanks a lot for your help!

You're welcome