# sequence formula

• Mar 19th 2008, 11:09 PM
jzellt
sequence formula
I need find a closed form for this recurrence relation:

b sub n = b sub n-1 + n^2 for n >0 and b sub 0 = 3

Anybody know how to do this?
• Mar 19th 2008, 11:28 PM
earboth
Quote:

Originally Posted by jzellt
I need find a closed form for this recurrence relation:

b sub n = b sub n-1 + n^2 for n >0 and b sub 0 = 3

Anybody know how to do this?

There is probably a more elegant way than this:

1. Calculate the first five values
2. Calculate the differences between consecutive values
3. Repeat this process until you get a constant sequence.
4. Determine the number of repetitions. That is the degree of the closed form:
Code:

degree 3:  3    4    8    17    33    ... degree 2:    1    4    9    16    ... degree 1:      3    5    7    ... degree 0:        2    2
That means you are looking for a term with degree 3:

$\displaystyle s_n = a\cdot n^3+b\cdot n^2+c\cdot n + d$

From the initial row you know:
s_0 = 3
s_1 = 4
s_2 = 8
s_3 = 17

Thus you have to solve a system of simultaneous equations:
$\displaystyle \left|\begin{array}{r}d=3\\a+b+c+d = 4\\8a+4b+2c+d=8\\27a+9b+3c+d=17\end{array}\right.$

After a few steps you'll get:

$\displaystyle s_n = \frac13 n^3 + \frac12 n^2 + \frac16n+3$