Finding the nth term of a sequence (formula?)

[Sandie]

I am trying to help my son find the equations for finding the nth terms. We can figure out the pattern and the nth term, but he has to have an equation which is where we are getting stuck.

Example:

Term     1,      2,     3,     4,     5,     6,    ....     20

Value    4,     15,    32,    55,    84,   119,           ???
           \   /   \  /   \  /   \  /   \  /          
             11     17     23     29     35
               \  /   \  /    \  /   \  / 
                 6      6       6      6

We factored the bottom numbers to:

. . .2x2. . .3x5. . .4x8. . .5x11. . .6x14. . .7x17

We see that the first factor increases by 1 and the second factor increases by 3. We don't know how to get a formula for this. Any help is appreciated. When looking at the preview, the numbers don't line up properly. I'm not sure how else to explain it.

 

[tiaa]

i have tried to do it hope tht it helps.

SOLUTION:

LET THE TERM BE n AND THE VALUE BE v.

v= 3(n^2)+2n-1.

applying the values given on the formula:

v=3((1^2))+2(1)-1
v=4


v=3(2^2)+2(2)-1
v=15

v= 3(3^2)+2(3)-1
v=32

v=3(4^2)+2(4)-1
v=55

v=3(5^2)+2(5)-1
v=84

v=3(6^2)+2(6)-1
v=119


PS: n^2 stands for n squared.

 

[Denis]

do a http://www.google.com search on "introduction to geometric series"

 

[Sandie]

I see how the equation works, but how did you come up with the equation?

 

[galactus]

If I understand correctly, it may help to use finite differences.

Here's a link. It's not difficult, just rather tedious.

http://mathforum.org/library/drmath/view/53223.html

 

[TchrWill]

Re: Finding the nth term

I am trying to help my son find the equations for finding the nth terms. We can figure out the pattern and the nth term, but he has to have an equation which is where we are getting stuck....

Lay out the terms.

n....1....2....3....4....5....6....7
N...4...15..32...55..84..119.
Diff..11..17..23..29...35
Diff.....6....6....6....6

With the 2nd differences being constant, we have what is referred to as a finite difference series with the 2nd differences beiing constant at 6. The equation of the nth term is of the form an^2 + bn + c.

Using the data, we hace

a(1^2) + b(1) + c = 4 or a + b + c = 4
a(2^2) + b(2) + c = 15 or 4a + 2b + c = 15
a(3^2) + b(3) +c = 32 or 9a + 3b + c =32

Solving, a = 3, b = 5 and c = 3

Therefore, we have the nth term Nn = 3n^2 + 2n - 1 = (3n + 1)(n - 1).

 

[soroban]

Re: Finding the nth term

Hello, Sandie!

You may be expected to be be familiar with the theory for this problem.
There are a number of approaches . . . Here's one of them.

      n        1       2       3       4       5       6

    f(n)       4       15      32      55      84     119
                 \   /   \   /   \   /   \   /   \   /
  1st diff         11      17      23      29      35
                     \   /   \   /   \   /   \   /
  2nd diff             6       6       6       6

We get a row of constants at the second differences.
Hence, the generating function, \(\displaystyle f(n)\), is of the second degree (a quadratic).

Then the function is of the form: \(\displaystyle \,f(n)\:=\:an^2\,+\,bn\,+\,c\)
\(\displaystyle \;\;\)and we must determine \(\displaystyle a,\;b,\;c.\)


We have three "unknowns", so we need three equations.
\(\displaystyle \;\;\)Use the first three values from our chart.

We know that:\(\displaystyle \;\begin{array}{ccc}f(1)\,=\,4\;\;\Rightarrow\;\;a\cdot1^2\,+\,b\cdot1\,+\,c\:=\:4\;\;\Rightarrow\;\;a\,+\,b\,+\,c\:=\:4 \\ f(2)\,=\,15\;\;\Rightarrow\;\;a\cdot2^2\,+\,b\cdot2\,+\,c\:=\:15 \;\;\Rightarrow\;\; 4a\,+\,2b\,+\,c\:=\:15 \\ f(3)\,=\,32\;\;\Rightarrow\;\;a\cdot3^2\,+\,b\cdot3\,+\,c\:=\:32\;\;\Rightarrow\;\;9a\,+\,3b\,+\,c\:=\:32 \end{array}\)

And solve the system of equations: \(\displaystyle \;\begin{array}{ccc}[1]\\[2]\\[3]\end{array}\;\begin{array}{ccc}a\,+\,b\,+\,c\:=\:4 \\ 4a\,+\,2b\,+\,c\:=\:15 \\ 9a\,+\,3b\,+\,c\:=\:32\end{array}\)

Subtract [1] from [2]: \(\displaystyle \:3a\,+\,b \:=\:11\;\;[4]\)
Subtract [2] from [3]: \(\displaystyle \:5a\,+\,b\:=\:17\;\;[5]\)

Subtract [4] from [5]: \(\displaystyle \:2a\,=\,6\;\;\Rightarrow\;\;a\,=\,3\)

Substitute into [4]: \(\displaystyle \:3\cdot3\,+\,b\:=\:11\;\;\Rightarrow\;\;b\,=\,2\)

Substitute into [1]: \(\displaystyle \:3\,+\,2\,+\,c\:=\:4\;\;\Rightarrow\;\;c\,=\,-1\)


Therefore: \(\displaystyle \:f(n)\;=\;3n^2\,+\,2n\,-\,1\;\;\) . . . ta-DAA!


[Edit: Too fast for me, TchrWill!]

 

[Sandie]

Thanks! I think we got it now! You guys are awesome.