# This Sum I am unable to solve

1. Jun 30, 2004

### Pattielli

Sum(from i=0 to N) of (i^4/2-3i^2/2+1)
Can you tell me how to calculate the value of that sum ?
Thank you,

- Pattielli

2. Jun 30, 2004

### matt grime

split into two smaller sums, factor out the constants and use the formulae for the sums of 4th and second powers that I haven't memorized but which should be googlable and that the constant term at the end contributes N+1 to the sum.

3. Jun 30, 2004

### AKG

$$\sum _{i=0} ^N \frac{1}{2}i^4 - \frac{3}{2}i^2 + 1$$

$$= 0.5\sum i^4 + 1.5 \sum i^2 + N + 1$$*

You need to find a general closed-form solution to the sum of the first N squares and first N $i^4$ (whatever you'd call them).

$$\sum _{i=0} ^N = \frac{N(N+1)(2N+1)}{6}$$

For the $i^4$, I don't have the solution memorized, nor can I think of a way right now to solve it.

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Check this link out. It's probably a good thing to learn, and it also gives the formula you need:

$$\sum _{i=0} ^n = (1/5)n^5 + (1/2)n^4 + (1/3)n^3 - (1/30)n$$

*Edited based on matt grime's observation

Last edited: Jun 30, 2004
4. Jun 30, 2004

### matt grime

there are N+1 terms in the sum so your last term, comig from the constant, is out by 1.

5. Jun 30, 2004

### Gokul43201

Staff Emeritus
Just to make this clear :

$$\sum _{i=0} ^N i^2= \frac{N(N+1)(2N+1)}{6}$$

The sum of fourth powers will be a polynomial of order 5. Use the values of the sum for N=1..5 to solve for the 5 coefficients (clearly, there's no constant term).