finding n

1) That makes little sense.
2) It is unlikely there is a unique answer.
3) What have you tried?
 
Hello, I jus dunno!

What is the original wording of the problem?

\(\displaystyle \begin{array}{c|cccccccc}n &1 & 2 & 3 & 4 & \hdots & 10 \\ \hline t_n & 2 & 3 & 0 \end{array}\)

What is the formula to find \(\displaystyle n\) ?
Click to expand...

It's a silly question, isn't it?
We already know the values of \(\displaystyle n\!:\;1,2,3\hdots 10\)
 
soroban said:
Hello, I jus dunno!

What is the original wording of the problem?


It's a silly question, isn't it?
We already know the values of \(\displaystyle n\!:\;1,2,3\hdots 10\)
Click to expand...


We were wondering what the pattern was from n to tn. What pattern is the same for all?
Ex.

n 1 2 3 4 5... 10
tn 2 4 6 8 10...20

therefore the nth= 2n+0
 
Hello, I jus dunno!

\(\displaystyle \begin{array}{c|cccc} n & 1 & 2 & 3 & \hdots \\ \hline t_n & 2 & 3 & 0 & \hdots \end{array}\)

What is the formula to for \(\displaystyle t_n\) ?
Click to expand...

Code: 
      |
      |         B
      |         o
      |   A   (2,3)
      |   o
      | (1,2)
      |
      |              C
  - - + -  - - - - - o - -
      |             (3,0)
There are zillions of graphs that pass through those three points.

The simplest would be a quadratic: .\(\displaystyle t(n) \:=\:-2t^2 + 7t - 3\)
 
soroban said:
Hello, I jus dunno!


Code: 
      |
      |         B
      |         o
      |   A   (2,3)
      |   o
      | (1,2)
      |
      |              C
  - - + -  - - - - - o - -
      |             (3,0)
There are zillions of graphs that pass through those three points.

The simplest would be a quadratic: .\(\displaystyle t(n) \:=\:-2t^2 + 7t - 3\)
Click to expand...


Hello and thank you! It does work, but how do you get those numbers such as -2, +7 and the -3.

We only learned very, very simple i.e.,

n 1 2 3
tn 3 7 11 and we got tn = 4n-1

Is there an answer for this next question?

n 1 2 3 4 5 6 .... 10
tn 1 5 7 9 ? ? ?

thank you, again
 
There is another simple possibility: t(n) = the remainder after division by 4 of (n+1). So t(4)=1.
 
JeffM said:
If you are given n points, there is a unique polynomial of degree (n - 1) that passes through those points. The coefficients of that polynomial can be found by solving a system of simultaneous linear equations. That process sounds more demanding than what I would expect in a first year algebra course.

It may be helpful if you state EXACTLY what the problem asks you to do. Your most recent problem can be solved by finding a cubic equation, but I somehow doubt that is what you are expected to do.
Click to expand...


Hello, we have explained in the prior posts that we are trying to find what the equation should be in order to fill in the rest of the numbers on the bottom. The answer that was given by another member was great, it is just that we do not know how he/she got -2, +7 and the -3. The last question is accurate and we cannot find the equation that works to answer what should be under the 4, the 5, the 6, etc.
 
I jus dunno said:
Hello, we have explained in the prior posts that we are trying to find what the equation should be in order to fill in the rest of the numbers on the bottom. The answer that was given by another member was great, it is just that we do not know how he/she got -2, +7 and the -3. The last question is accurate and we cannot find the equation that works to answer what should be under the 4, the 5, the 6, etc.
Click to expand...

It isn't so elementary to be considered "basic algebra" so I doubt that method is the one which is intended.

You are given 3 points: (1,2), (2,3), (3,0), so you are looking for a second degree polynomial which fits this data.

\(\displaystyle f(x)=ax^2+bx+c\)

Your goal is to find a, b, and c.

You need f(1) = 2, f(2) = 3, f(3) = 0. So you plug them in to get three equations:

\(\displaystyle a(1)^2 + b(1) + c = 2\)
\(\displaystyle a(2)^2 + b(2) + c = 3\)
\(\displaystyle a(3)^2 + b(3) + c = 0\)

Or

\(\displaystyle a + b + c = 2\)
\(\displaystyle 4a + 2b + c = 3\)
\(\displaystyle 9a + 3b + c = 0\)

Then you may solve the above using (1) the method of substitution, (2) the method of addition, or (3) Cramer's rule for linear equations. If you do not know what these are, then I would consult Wikipedia or an algebra book for a slightly more advanced algebra class.

This is not so hard to do in this case. But if you had been given 5 data points, it is essentially equivalent in length to computing the solution to the above question 12 times! Not so trivial.

I maintain that \(\displaystyle t_n = (n+1) - 4\lfloor\frac{n+1}{4}\rfloor\) is the simplest, where \(\displaystyle \lfloor y\rfloor\) is the floor function. This is the remainder function I specified in my last post.
 
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