## anonymous 5 years ago Hi, I need to prove Sn= 6n^2-6n+1. How can I d othis?

1. anonymous

$6n ^{2}-6n+1$ HOW CAN I PROVE THIS?

2. anonymous

3. anonymous

Just Math, an IB course. I don't know how to prove this... at all, any questions to lead me towards something productive?

4. anonymous

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

5. anonymous

not enough info o_o

6. anonymous

What is Sn?

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

Oh I see.

11. anonymous

haha now it's solvable ^^

12. anonymous

so how can I do this?

13. 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?

14. anonymous

But that's just checking, I need to know how to prove the equation without knowing the formula already..?

15. amistre64

either factor it or complete the square

16. anonymous

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

17. 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)

18. anonymous

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

19. anonymous

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

20. 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.

21. anonymous

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

22. 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?

23. 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.

24. 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))

25. anonymous

Now we are lost. You said you wanted to prove something. You didn't say you want to find an equation. We apologize.

26. 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: $f(n)=1+12({1 \over 2}n(n-1))=1+6(n(n-1))=1+6n^2-6n$

27. anonymous

I hope that helps.

28. anonymous

THANK YOU SO MUCH !

29. anonymous

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

30. anonymous

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

31. anonymous

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