Hi, I need to prove Sn= 6n^2-6n+1. How can I d othis?

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Hi, I need to prove Sn= 6n^2-6n+1. How can I d othis?

Mathematics
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\[6n ^{2}-6n+1\] HOW CAN I PROVE THIS?
Need more info, what subject or course are you doing?
Just Math, an IB course. I don't know how to prove this... at all, any questions to lead me towards something productive?

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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?
get any point yu have and put on your equantion: f(5) = 121 121 = 6(5)^2 - 6(5) + 1 121 = 150 - 30 + 1 121 = 121 (True) is that right?
But that's just checking, I need to know how to prove the equation without knowing the formula already..?
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
Look above at the example of Mathmind, that is what you do for each term. You just dismissed it as that's just checking. That is what you are supposed to do, check to see if it is true or not.
so how would I prove the equation if I don't know what the equation is to begin with...
You were given the info. You gave us the info. I think you are a little lost, but that is OK. When they say f(1)=1. They are saying that 6(1)^2-6(1)+1=1 I this true or not?
Yeah I know that's true. But I'm not proving that the points match, I'm using the points to find an equation, say that I don't know what the equation is.
f(1)=1 f(2)=1+12 f(3)=1+12+2(12)=1+12(1+2) f(4)=1+12+2(12)+3(12)=1+12(1+2+3) f(n)=1+12(1+2+3+..(n-1))
Now we are lost. You said you wanted to prove something. You didn't say you want to find an equation. We apologize.
There is a formula for the summation (1+2+3+..+(n-1)), that's: \[1+2+3+...+(n-1)={1 \over 2}n(n-1)\] Plug that in the previous equation, you get: \[f(n)=1+12({1 \over 2}n(n-1))=1+6(n(n-1))=1+6n^2-6n\]
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.

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