how do you think is this gonna prove the question.How did you get
[imath]1 - n - n^2 - n^3 - n^4 - n^5 - n^6 = \dfrac{1 - n^6}{1 + n}[/imath]
That isn't true for all n!
-Dan
Addendum: It is true that the sum you listed is [imath]1 - \dfrac{n^6 - 1}{n - 1}[/imath].
Neither can I. I was simply pointing out a mistake.how do you think is this gonna prove the question.
[imath](1 - \dfrac{x^6 - 1}{x - 1})^n[/imath]
I am still cannot find good relation between the problem and its generating function.
me neither, even the source question was hard to understand for me.I cannot even figure out what you are asking: “number of way in which n achieved by rolling the dice any time” makes no sense in English.
thanks, the dice problem is a known subject and if we wanna look at the other ways of showing generating function :Wherever you got this problem or whoever wrote the problem is mistaken.
If a die (standard cubic die, numbered 1 to six) is tossed \[imath]n[/imath] times then
the generating function [imath](x+x^2+x^3+x^4+x^5+x^6)^n[/imath] tells us the number of ways a sum of the faces can be made. SEE THIS
For example a gamming die is tossed eight them ( {\sum\limits_{k = 1}^6 {{x^k}} } \right)^8}[/imath] [imath]{\leftcontains the term [imath]3368x^{41}[/imath]. Which tell s that the sum [imath]41[/imath] can be gotten in [imath]3368[/imath] ways so the probability is [imath]\dfrac{3368}{6^8}[/imath]. We also use that sum if we roll [imath]n[/imath] dice.