Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

not enough info o_o

Answer 6

What is Sn?

Answer 7

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

Answer 8

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

Answer 9

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

a quadratic equation.

Answer 10

Oh I see.

Answer 11

haha now it's solvable ^^

Answer 12

so how can I do this?

Answer 13

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?

Answer 14

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

Answer 15

either factor it or complete the square

Answer 16

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

Answer 17

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)

Answer 18

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

Answer 19

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

Answer 20

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.

Answer 21

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

Answer 22

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?

Answer 23

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.

Answer 24

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))

Answer 25

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

Answer 26

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:
\[f(n)=1+12({1 \over 2}n(n-1))=1+6(n(n-1))=1+6n^2-6n\]

Answer 27

I hope that helps.

Answer 28

THANK YOU SO MUCH !

Answer 29

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

Answer 30

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

Answer 31

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