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\[6n ^{2}-6n+1\] HOW CAN I PROVE THIS?

Need more info, what subject or course are you doing?

6n^2-6n+1 = 0 is that what yu are trying to prove?

not enough info o_o

What is Sn?

f(1) = 1
f(2) = 13
f(3) = 37
f(4) = 73
f(5) = 121
f(6) = 181

It's actually, f(n) = 6n^2 - 6 + 1

wait, ANOTHER TYPO.
f(n) = 6n^2 - 6n + 1
a quadratic equation.

Oh I see.

haha now it's solvable ^^

so how can I do this?

either factor it or complete the square

i think she has to find the formula with those points she has

1st term=12+1
2nd term=13+2(12)
3rd term=37+3(12)
4th term=73+4(12)
.
.
nth term=(n-1)th term+n(12)

Oh sorry, first term is 1. So, the formula should be:
\[a_n=a_{n-1}+12(n-1)\]

and then.. ? what do I do from there to get 6n^2 - 6n + 1

so how would I prove the equation if I don't know what the equation is to begin with...

I hope that helps.

THANK YOU SO MUCH !

Can you help me explain this? I'm not sure how I would explain this

The steps are all clear. Try to read them again, the last two comments before "I hope that hels".

If there is something you don't understand, let me know.