Quote:
Originally Posted by wpolly
(120 + 94 n + 5 n^2 + 25 n^3  5 n^4 + n^5)/120

Thanks to your response to this problem in two parts.
I think they should be upgraded to the math forum as they contain deep math concepts.
Kindly note Xyzzy if you read this.
Thrillar:. A good spirited attempt but your formula does not agree with the 6th term onwards. Yes every time you fall Get UP and don’t remain fallen!
Cheesehead: A good explanation but not rigorously correct. The spots on the circle formed by concurrent lines are not to be confused with points formed by intersecting lines which tho’ of zero space are to be counted. The spots are not included in the summation.
Wpolly: You are on the right track. You are out by a factor of appx. 5n, which if you divide by it, your answer will be nearly correct but not exactly. I’m sure if you rework your calculations you should have the right answer. If you recall in ‘Circle 1’ I had mentioned that the function is a quartic and not the quintic you have given.
Now to give a further hint I give the next three terms after 6th term T6 = 31.
These are T7 =57; T8 =99 ; T9 ==163.
I give two methods:
1) Use the method of finite differences and the first 8 terms are sufficient to evaluate the function.
2)
You may also use Pascals triangle and make an appropriate diagonal cut from the 6th line downwards. The summation of each line gives a term. From the 6th line omit the no. 1 on the R.H.S.. On the 7th. line omit two no.s. viz 6 and ! on the R.H.S and form the diagonal downwards omitting the terms on the R.H.S. of the diagonal.
[R.H.S. =Right hand side.]
Try it out as it’s quite easy.
If you have understood any of these 2 methods (the second is easier and straight forward) you should be able to predict the tenth term.
T10 = ?
I await your answers after which I will work out the quartic in full and give a concise formula how to remember it in only 3 terms given by an American math’ cian (Name concealed ) who worked it out. .
Mally