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?

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- Mar 19th 2008, 11:09 PMjzelltsequence 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 PMearboth
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

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

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:

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