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How to find

$$\sum\limits_{i=1,i\neq 5,12,23,45}^{100} i^3$$ One way I know is $$\sum\limits_{i=1}^{100} i^3-5^3-12^3-23^3-45^3$$ But when the missing terms in the sequence become large it is difficult and time taking to find the sum. Is there any way to calculate this type of summation.

Litun
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    If there is a pattern in the non-ommited terms, then you can rewrite your sum so that it only takes those terms. Otherwise there is no magic formula for that: you either subtract the superfluous terms, or you separate you summation so that it does not include them. – Olivier Jan 16 '14 at 07:05
  • @Oliver is there any mistake in my answer/ – dato datuashvili Jan 16 '14 at 07:05
  • @Litun welcome to math site,your problem is clear and +1,but you may search a bit to find formula for cubic sum,which i have posted – dato datuashvili Jan 16 '14 at 07:07
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    There is no mistake. – Olivier Jan 16 '14 at 07:08

2 Answers2

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you solution is following one

$$\sum\limits_{i=1}^{100} i^3-5^3-12^3-23^3-45^3$$

now you may calculate what is cubic sum of integers from $1$ to $100$ with basic definition of formula

The sum of the first $k$ cubic numbers is equal to

enter image description here

please look there http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/integration/sums.html

dato datuashvili
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Let us make the problem more general in the sense that the summation will be from i=1 to i=N and that you want to exclude terms following the kind of pattern you gave.

For the summation, just use the classical formula given by dato datuashvili.

Now, the numbers you exclude in your problem follow the general equation

j(i) = - 5 + 83 i / 6 - 5 i^2 + 7 i^3 / 6

You could easily check that, whatever "i" could be, j(i) is an integer.

So, from the total summation, you must exclude the sum of the cubes of j(i) for i running from 1 to p. If you expand the cube of j(i), you end with sums of i^n (n running from 1 to 9) and i from 1 to p. Each of this sums as a closed form; it is even possible to directly establish for your case that the sum of the cubes of excluded terms is given by

p (-2246400 - 2820636 p + 10899980 p^2 + 4880645 p^3 - 6655740 p^4+ 4947222 p^5 -
- 1908660 p^6 + 543165 p^7 - 89180 p^8 + 9604 p^9) / 60480

Applied to your problem (p=4), this gives for the sum of the excluded terms the value of 105145 which is correct.

Claude Leibovici
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  • If the excluded terms don't follow any pattern, then what will the approach to get the sum other than the method what dato datuashvili has explained – Litun Jan 16 '14 at 10:13
  • @Litun. You cannot find anything simpler than what dato datuashvili proposed. I even suppose that this is the only way to do it .... except if you want to compute the sum adding all terms. What was given by dato datuashvili is a remarkable identity. – Claude Leibovici Jan 16 '14 at 10:19
  • Ok, How to get that particular pattern if there is any? using trial method or any other technique is there. – Litun Jan 16 '14 at 10:24
  • @Litun. Start plotting the excluded values. In your case, you have four points (x,y), namely (1,5), (2,12), (3,23), (4,45). You can also search for completing the sequence (http://alteredqualia.com/visualization/hn/sequence/). In your case, the sequence continue with 85, 150, 247. Try to fit with a polynomial; if the fit is perfect, you are done. If it is not, just use your calculator. – Claude Leibovici Jan 16 '14 at 10:32