Originally Posted by

**math-** hi everyone!

ok, i have an investigation where i must consider some fraction sequences.

first row- 1/2, 2/3, 3/4, 4/5 5/6, 6/7, 78

second row- 1/6, 1/12, 1/20, 1/30, 1/42, 1/56

third row- 1/12, 1/30, 1/60, 1/105, 1/168

fourth- 1/20, 1/60, 1/140, 1/280

...

Hello,

as Geometor suggested I'll use the denominators only:

Consider the following sequence and calculate the diffferences between the numbers: Code:

6 12 20 30 42 56
6 8 10 12 14
2 2 2 2

The differences of the differences yield a constant sequence. Thus your sequence is of the second grade/degree(?) and the general equation of such a sequence is:

$\displaystyle s(n)=a \cdot n^2 + b \cdot n + c$

You know

$\displaystyle

\begin{array}{l}s(1) = 6\\s(2)=12\\s(3)=20\end{array}~\implies~ \left|\begin{array}{lcr}a+b+c&=&6\\ 4a+2b+c&=&12\\9a+3b+c&=&20\end{array}\right.$

Solve this system of simultaneous equations. I've got: a = 1, b = 3, c = 2

That means the sequence have the equation

$\displaystyle s(n) = n^2 + 3n + 2$

And the numbers in the second row can be calculated by:

$\displaystyle r_2(n) = \frac1{n^2 + 3n + 2}$

If you apply this method to your 3rd row you'll get a sequence of third grade/degree(?). Thus you have to solve a system of 4 simultaneous equations:

$\displaystyle

\begin{array}{l}s(1) = 12\\s(2)=30\\s(3)=60\\s(4) = 105\end{array} ~\implies~ \left|\begin{array}{lcr}a+b+c+d&=&12\\ 8a+4b+2c+d&=&30\\27a+9b+3c+d&=&60\\ 64a+16b+4c+d&=&105\end{array}\right.$

Solve this system. I've got as the final result that the numbers in the third row can be calculated by:

$\displaystyle r_3(n) = \frac1{\frac12 \cdot n^3 + 3n^2 + \frac{11}2 \cdot n +3}$