This Sum I am unable to solve

[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

[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.

[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.

-----------------------------------

Check this link (http://mathforum.org/library/drmath/view/56383.html) 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

[matt grime]

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

[Gokul43201]

\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.



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).

Or Google harder.

[mathwonk]

gokul 43201 gave the quickest solution method, via lagrange interpolation.

there is another recursive method, for findign formulas for the first n terms of any series of rth powers of integers, described in a footnote to the courant calculus book vol 1, in the first few pages. when i saw that i knew I had found a real math book, as there was more content in that footnote than in all math books i had used previously.