anonymous
  • anonymous
Hi, I need to prove Sn= 6n^2-6n+1. How can I d othis?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[6n ^{2}-6n+1\] HOW CAN I PROVE THIS?
anonymous
  • anonymous
Need more info, what subject or course are you doing?
anonymous
  • anonymous
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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anonymous
  • anonymous
6n^2-6n+1 = 0 is that what yu are trying to prove?
anonymous
  • anonymous
not enough info o_o
anonymous
  • anonymous
What is Sn?
anonymous
  • anonymous
f(1) = 1 f(2) = 13 f(3) = 37 f(4) = 73 f(5) = 121 f(6) = 181
anonymous
  • anonymous
It's actually, f(n) = 6n^2 - 6 + 1
anonymous
  • anonymous
wait, ANOTHER TYPO. f(n) = 6n^2 - 6n + 1 a quadratic equation.
anonymous
  • anonymous
Oh I see.
anonymous
  • anonymous
haha now it's solvable ^^
anonymous
  • anonymous
so how can I do this?
anonymous
  • anonymous
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?
anonymous
  • anonymous
But that's just checking, I need to know how to prove the equation without knowing the formula already..?
amistre64
  • amistre64
either factor it or complete the square
anonymous
  • anonymous
i think she has to find the formula with those points she has
anonymous
  • anonymous
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)
anonymous
  • anonymous
Oh sorry, first term is 1. So, the formula should be: \[a_n=a_{n-1}+12(n-1)\]
anonymous
  • anonymous
and then.. ? what do I do from there to get 6n^2 - 6n + 1
anonymous
  • anonymous
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.
anonymous
  • anonymous
so how would I prove the equation if I don't know what the equation is to begin with...
anonymous
  • anonymous
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?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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))
anonymous
  • anonymous
Now we are lost. You said you wanted to prove something. You didn't say you want to find an equation. We apologize.
anonymous
  • anonymous
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: \[f(n)=1+12({1 \over 2}n(n-1))=1+6(n(n-1))=1+6n^2-6n\]
anonymous
  • anonymous
I hope that helps.
anonymous
  • anonymous
THANK YOU SO MUCH !
anonymous
  • anonymous
Can you help me explain this? I'm not sure how I would explain this
anonymous
  • anonymous
The steps are all clear. Try to read them again, the last two comments before "I hope that hels".
anonymous
  • anonymous
If there is something you don't understand, let me know.

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